Approximately 95% of housing prices are between a low price of $164,000 and a high price of $192,000.
To determine the range of housing prices between which approximately 95% of prices fall, we can use the empirical rule, also known as the 68-95-99.7 rule. According to this rule, for a normal distribution:
- Approximately 68% of the data falls within one standard deviation of the mean.
- Approximately 95% of the data falls within two standard deviations of the mean.
- Approximately 99.7% of the data falls within three standard deviations of the mean.
In this case, the mean housing price is $178,000, and the standard deviation is $7,000. To find the low and high prices within which approximately 95% of the housing prices fall, we can apply the empirical rule.
First, we calculate one standard deviation:
Standard deviation = $7,000
Next, we calculate two standard deviations:
Two standard deviations = 2 * $7,000 = $14,000
To find the low price, we subtract two standard deviations from the mean:
Low price = $178,000 - $14,000 = $164,000
To find the high price, we add two standard deviations to the mean:
High price = $178,000 + $14,000 = $192,000
Therefore, approximately 95% of housing prices are between a low price of $164,000 and a high price of $192,000.
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1. Find the general solution for each of the following differential equations (10 points each). a. y" +36y=0 b. y"-7y+12y=0
a. For the differential equation y" + 36y = 0, assume y = [tex]e^(rt)[/tex]. Substituting it in the equation yields r² + 36 = 0, giving imaginary roots r = ±6i. The general solution is y = Acos(6x) + Bsin(6x).
b. For the differential equation y" - 7y + 12y = 0, assume y = [tex]e^(rt)[/tex]. Substituting it in the equation yields r² - 7r + 12 = 0, giving roots r = 3 or r = 4. The general solution is y = [tex]C1e^(3x) + C2e^(4x)[/tex].
The detailed calculation step by step for each differential equation:
a. y" + 36y = 0
Assume a solution of the form y = e^(rt), where r is a constant.
1. Substitute the solution into the differential equation:
y" + 36y = 0
[tex](e^(rt))" + 36e^(rt)[/tex]= 0
2. Take the derivatives:
[tex]r^2e^(rt) + 36e^(rt)[/tex]= 0
3. Factor out [tex]e^(rt)[/tex]:
[tex]e^(rt)(r^2 + 36)[/tex]= 0
4. Set each factor equal to zero:
[tex]e^(rt)[/tex] = 0 (which is not possible, so we disregard it)
r² + 36 = 0
5. Solve the quadratic equation for r²:
r² = -36
6. Take the square root of both sides:
r = ±√(-36)
r = ±6i
7. Rewrite the general solution using Euler's formula:
Since [tex]e^(ix)[/tex] = cos(x) + isin(x), we can rewrite the general solution as:
y = [tex]C1e^(6ix) + C2e^(-6ix)[/tex]
= C1(cos(6x) + isin(6x)) + C2(cos(6x) - isin(6x))
= (C1 + C2)cos(6x) + i(C1 - C2)sin(6x)
8. Combine the arbitrary constants:
Since C1 and C2 are arbitrary constants, we can combine them into a single constant, A = C1 + C2, and rewrite the general solution as:
y = Acos(6x) + Bsin(6x), where A and B are arbitrary constants.
b. y" - 7y + 12y = 0
Assume a solution of the form y = [tex]e^(rt)[/tex], where r is a constant.
1. Substitute the solution into the differential equation:
y" - 7y + 12y = 0
[tex](e^(rt))" - 7e^(rt) + 12e^(rt)[/tex]= 0
2. Take the derivatives:
[tex]r^2e^(rt) - 7e^(rt) + 12e^(rt)[/tex]= 0
3. Factor out [tex]e^(rt)[/tex]:
[tex]e^(rt)(r^2 - 7r + 12)[/tex] = 0
4. Set each factor equal to zero:
[tex]e^(rt)[/tex] = 0 (which is not possible, so we disregard it)
r² - 7r + 12 = 0
5. Factorize the quadratic equation:
(r - 3)(r - 4) = 0
6. Solve for r:
r = 3 or r = 4
7. Write the general solution:
The general solution for the differential equation is:
y =[tex]C1e^(3x) + C2e^(4x)[/tex]
Alternatively, we can rewrite the general solution using the exponential form of complex numbers:
y = [tex]C1e^(3x) + C2e^(4x)[/tex]
where C1 and C2 are arbitrary constants.
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A tank initially contains 10 gal of fresh water. At t = 0, a brine solution containing 0.5 Ib of salt per gallon is poured into the tank at the rate of 2 gal/min, while the well-stirred mixture leaves the tank at the same rate. find (a) the amount and (b) the concentration of salt in the tank at any time t.
(a) The amount of salt in the tank at any time t can be calculated by considering the rate at which the brine solution is poured in and the rate at which the mixture leaves the tank.
(a) To find the amount of salt in the tank at any time t, we need to consider the rate at which the brine solution is poured in and the rate at which the mixture leaves the tank.
The rate at which the brine solution is poured into the tank is 2 gal/min, and the concentration of salt in the solution is 0.5 lb/gal. Therefore, the rate of salt input into the tank is 2 gal/min * 0.5 lb/gal = 1 lb/min.
At the same time, the mixture is leaving the tank at a rate of 2 gal/min. Since the tank is well-stirred, the concentration of salt in the mixture leaving the tank is assumed to be uniform and equal to the concentration of salt in the tank at that time.
Hence, the rate at which salt is leaving the tank is given by the concentration of salt in the tank at time t multiplied by the rate of outflow, which is 2 gal/min.
The net rate of change of salt in the tank is the difference between the rate of input and the rate of output:
Net rate of change = Rate of input - Rate of output
= 1 lb/min - (2 gal/min * concentration of salt in the tank)
Since the volume of the tank remains constant at 10 gal, the rate of change of salt in the tank can be expressed as the derivative of the amount of salt with respect to time:
dy/dt = 1 lb/min - 2 * concentration of salt in the tank
This is a first-order linear ordinary differential equation that we can solve to find the amount of salt in the tank at any time t.
(b) The concentration of salt in the tank at any time t can be found by dividing the amount of salt in the tank by the volume of water in the tank.
Concentration = Amount of salt / Volume of water in the tank
= y(t) / 10 gal
By substituting the solution for y(t) obtained from solving the differential equation, we can determine the concentration of salt in the tank at any time t.
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A solid, G is bounded in the first octant by the cylinder x^2 +z^2 =3^2, plane y=x, and y=0. Express the triple integral ∭ G dV in four different orientations in Cartesian coordinates dzdydx,dzdxdy,dydzdx, and dydxdz. Choose one of the orientations to evaluate the integral.
The value of the triple integral is -27 when expressed in the dzdydx orientation.
Given, a solid, G is bounded in the first octant by the cylinder x²+z²=3², plane y=x, and y=0.
We are to express the triple integral ∭ G dV in four different orientations in Cartesian coordinates dzdydx, dzdxdy, dydzdx, and dydxdz and choose one of the orientations to evaluate the integral.
In order to express the triple integral ∭ G dV in four different orientations, we need to identify the bounds of integration with respect to x, y and z.
Since the solid is bounded in the first octant, we have:
0 ≤ y ≤ x
0 ≤ x ≤ 3
0 ≤ z ≤ √(9 - x²)
Now, let's express the integral in each of the given orientations:
dzdydx: ∫[0,3] ∫[0,x] ∫[0,√(9 - x²)] dzdydx
dzdxdy: ∫[0,3] ∫[0,√(9 - x²)] ∫[0,x] dzdxdy
dydzdx: ∫[0,3] ∫[0,x] ∫[0,√(9 - x²)] dydzdx
dydxdz: ∫[0,3] ∫[0,√(9 - x²)] ∫[0,x] dydxdz
Let's evaluate the integral in the dzdydx orientation:
∫[0,3] ∫[0,x] ∫[0,√(9 - x²)] dzdydx
= ∫[0,3] ∫[0,x] [√(9 - x²)] dydx
= ∫[0,3] [(1/2)(9 - x²)^(3/2)] dx
= [-(1/2)(9 - x²)^(5/2)] from 0 to 3
= 27/2 - 81/2
= -27
Therefore, the value of the triple integral is -27 when expressed in the dzdydx orientation.
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Mention whether the following statements are true or false without giving any reasons. Assume that the functions ƒ : R → R and g : R → R are arbitrary functions.
(a) [1 point] ƒ ° ƒ = ƒ.
(b) [1 point] fog = gof.
(c) [1 point] ƒ and g are both one-to-one correspondences implies that ƒ o g and go f are both one-to-one correspondences.
(d) [1 point] ƒ and g are both onto does not imply that ƒ og and go ƒ are both onto. (e) [1 point] ƒ and g are both one-to-one implies that fog and go f are both one-to-one.
(f) [1 point] If ƒ o g is the identity function, then ƒ and g are one-to-one correspon- dences.
(g) [1 point] Suppose ƒ-¹ exists. Then ƒ-¹ need not be an onto function.
(h) [1 point] The size of the set of all multiples of 6 is less than the size of the set of all multiples of 3.
(i) [1 point] The size of the set of rational numbers is the same as the size of the set of real numbers in the range [0, 0.0000001].
(j) [1 point] The size of the set of real numbers in the range [1, 2] is the same or larger than the size of the set of real numbers in the range [1, 4].
The false statements arise from counterexamples or violations of these properties.
Is the integral of a continuous function always continuous?In this set of statements, we are asked to determine whether each statement is true or false without providing reasons.
These statements involve properties of functions and the sizes of different sets.
To fully explain the reasoning behind each statement's truth or falsehood, we would need to consider various concepts from set theory and function properties.
However, in summary, the true statements are based on established properties of functions and sets, such as composition, injectivity, surjectivity, and set cardinality.
Overall, a comprehensive explanation of each statement would require a more detailed analysis of the underlying concepts and properties involved.
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Let f (x) = (x+2)(3x-5)/(x+5)(2x – 1)
For this function, identify
1) the y intercept
2) the x intercept(s)
3) the Vertical asymptote(s) at x =
1) The y-intercept is (0, 2/5).
2) The x-intercepts are (-2, 0) and (5/3, 0).
3) The vertical asymptotes occur at x = -5 and x = 1/2.
How to identify the Y-intercept of function?1) To identify the properties of the function f(x) = (x+2)(3x-5)/(x+5)(2x-1):
To find the y-intercept, we set x = 0 and evaluate the function:
f(0) = (0+2)(3(0)-5)/(0+5)(2(0)-1) = (-10)/(5(-1)) = 2/5
Therefore, the y-intercept is at the point (0, 2/5).
How to identify the X-intercepts of function?2) To find the x-intercepts, we set f(x) = 0 and solve for x:
(x+2)(3x-5) = 0
From this equation, we can solve for x by setting each factor equal to zero:
x+2 = 0 --> x = -2
3x-5 = 0 --> x = 5/3
Therefore, the x-intercepts are at the points (-2, 0) and (5/3, 0).
How to identify the Vertical asymptotes of function?3) Vertical asymptotes occur when the denominator of a rational function equals zero. In this case, the denominator is (x+5)(2x-1), so we set it equal to zero and solve for x:
x + 5 = 0 --> x = -5
2x - 1 = 0 --> x = 1/2
Therefore, the vertical asymptotes occur at x = -5 and x = 1/2.
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Sketch the plane curve defined by the given parametric equations and find a corresponding x−y equation for the curve. x=−3+8t
y=7t
y= ___x+___
The x-y equation for the curve is y = (7/8)x + 2.625.
The given parametric equations are:
x = -3 + 8t
y = 7t
To find the corresponding x-y equation for the curve, we can eliminate the parameter t by isolating t in one of the equations and substituting it into the other equation.
From the equation y = 7t, we can isolate t:
t = y/7
Substituting this value of t into the equation for x, we get:
x = -3 + 8(y/7)
Simplifying further:
x = -3 + (8/7)y
x = (8/7)y - 3
Therefore, the corresponding x-y equation for the curve is:
y = (7/8)x + 21/8
In slope-intercept form, the equation is:
y = (7/8)x + 2.625
So, the x-y equation for the curve is y = (7/8)x + 2.625.
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Q 2: 9 points Give a regular expression for each of the following regular languages. You may use \( + \) and exponents as shorthand, but you clearly can't use the \( \cap \) and - operations. a) The s
Let's assume that the language in part (a) is intended to be "the set of strings that start with 's'." In that case, the regular expression for this language can be expressed as: The regular expression "s.*" matches any string that starts with the letter 's' followed by zero or more occurrences of any character (denoted by the '.' symbol).
The asterisk (*) indicates zero or more repetitions of the preceding character or group. Please note that this is just one example of a regular expression based on an assumption of the incomplete language description. If you intended a different language or have more specific requirements, please provide additional details, and I will be glad to assist you further.
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Complete the following items. For multiple choice items, write the letter of the correct response on your paper. For all other items, show or explain your work.Let f(x)=4/{x-1} ,
c. How are the domain and range of f and f⁻¹ related?
The domain of f is all real numbers except 1, and the range is all real numbers except 0. The domain and range of f⁻¹ are interchanged.
The function f(x) = 4/(x-1) has a restricted domain due to the denominator (x-1). For any value of x, the function is undefined when x-1 equals zero because division by zero is not defined. Therefore, the domain of f is all real numbers except 1.
In terms of the range of f, we consider the behavior of the function as x approaches positive infinity and negative infinity. As x approaches positive infinity, the value of f(x) approaches 0. As x approaches negative infinity, the value of f(x) approaches 0 as well. Therefore, the range of f is all real numbers except 0.
Now, let's consider the inverse function f⁻¹(x). The inverse function is obtained by swapping the x and y variables and solving for y. In this case, we have y = 4/(x-1). To find the inverse, we solve for x.
By interchanging x and y, we get x = 4/(y-1). Rearranging the equation to solve for y, we have (y-1) = 4/x. Now, we isolate y by multiplying both sides by x and then adding 1 to both sides:
yx - x = 4
yx = x + 4
y = (x + 4)/x
From this equation, we can see that the domain of f⁻¹ is all real numbers except 0 (since division by 0 is undefined), and the range of f⁻¹ is all real numbers except 1 (since the denominator cannot be equal to 1).
Therefore, the domain and range of f and f⁻¹ are interchanged. The domain of f becomes the range of f⁻¹, and the range of f becomes the domain of f⁻¹.
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A tower that is 35 m tall is to have to support two wires and start out with stability both will be attached to the top of the tower it will be attached to the ground 12 m from the base of each wire wires in the show 5 m to complete each attachment how much wire is needed to make the support of the two wires
The 34 m of wire that is needed to support the two wires is the overall length.
Given, a tower that is 35 m tall and is to have to support two wires. Both the wires will be attached to the top of the tower and it will be attached to the ground 12 m from the base of each wire. Wires in the show 5 m to complete each attachment. We need to find how much wire is needed to make support the two wires.
Distance of ground from the tower = 12 lengths of wire used for attachment of wire = 5 mWire required to attach the wire to the top of the tower and to ground = 5 + 12 = 17 m
Wire required for both the wires = 2 × 17 = 34 m length of the tower = 35 therefore, the total length of wire required to make the support of the two wires is 34 m.
What we are given?
We are given the height of the tower and are asked to find the total length of wire required to make support the two wires.
What is the formula?
Wire required to attach the wire to the top of the tower and to ground = 5 + 12 = 17 mWire required for both the wires = 2 × 17 = 34 m
What is the solution?
The total length of wire required to make support the two wires is 34 m.
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Find y as a function of x if x^2y′′+6xy′−14y=x^3
y(1)=3. V′(1)=3
y= _________
Answer: It is stated down below
Step-by-step explanation:
To solve the given second-order linear homogeneous differential equation, we can use the method of undetermined coefficients. Let's solve it step by step:
The given differential equation is:
x^2y'' + 6xy' - 14y = x^3
We assume a particular solution of the form y_p(x) = Ax^3, where A is a constant to be determined.
Now, let's find the first and second derivatives of y_p(x):
y_p'(x) = 3Ax^2
y_p''(x) = 6Ax
Substituting these derivatives back into the differential equation:
x^2(6Ax) + 6x(3Ax^2) - 14(Ax^3) = x^3
Simplifying the equation:
6Ax^3 + 18Ax^3 - 14Ax^3 = x^3
10Ax^3 = x^3
Now, comparing the coefficients on both sides of the equation:
10A = 1
A = 1/10
So, the particular solution is y_p(x) = (1/10)x^3.
To find the general solution, we need to consider the complementary solution to the homogeneous equation, which satisfies the equation:
x^2y'' + 6xy' - 14y = 0
We can solve this homogeneous equation by assuming a solution of the form y_c(x) = x^r, where r is a constant to be determined.
Differentiating y_c(x) twice:
y_c'(x) = rx^(r-1)
y_c''(x) = r(r-1)x^(r-2)
Substituting these derivatives back into the homogeneous equation:
x^2(r(r-1)x^(r-2)) + 6x(rx^(r-1)) - 14x^r = 0
Simplifying the equation:
r(r-1)x^r + 6rx^r - 14x^r = 0
(r^2 - r + 6r - 14)x^r = 0
(r^2 + 5r - 14)x^r = 0
For this equation to hold for all values of x, the coefficient (r^2 + 5r - 14) must be equal to zero. So we solve:
r^2 + 5r - 14 = 0
Factoring the equation:
(r + 7)(r - 2) = 0
This gives two possible values for r:
r_1 = -7
r_2 = 2
Therefore, the complementary solution is y_c(x) = C_1x^(-7) + C_2x^2, where C_1 and C_2 are constants.
The general solution is given by the sum of the particular and complementary solutions:
y(x) = y_p(x) + y_c(x)
= (1/10)x^3 + C_1x^(-7) + C_2x^2
To find the values of C_1 and C_2, we can use the initial conditions:
y(1) = 3
y'(1) = 3
Substituting these values into the general solution:
3 = (1/10)(1)^3 + C_1(1)^(-7) + C_2(1)^2
3 = 1/10 + C_1 + C_2
3 = 1/10 + C_1 + C_2 (Equation 1)
3 = (3/10) + C_1 + 1(C_2) (Equation 2)
From Equation 1, we get:
C_1 + C_2 = 3 - 1/10
From Equation 2, we get:
C_1 + C_2 = 3 - 3/10
Combining the equations:
C_1 + C_2 = 27/10 - 3/10
C_1 + C_2 = 24/10
C_1 + C_2 = 12/5
Since C_1 + C_2 is a constant, we can represent it as another constant, let's call it C.
C_1 + C_2 = C
Therefore, the general solution can be written as:
y(x) = (1/10)x^3 + C_1x^(-7) + C_2x^2
= (1/10)x^3 + Cx^(-7) + Cx^2
Thus, y as a function of x is given by:
y(x) = (1/10)x^3 + Cx^(-7) + Cx^2, where C is a constant.
Find a basis for the eigenspace corresponding to each listed eigenvalue of A
To find a basis for the eigenspace corresponding to each listed eigenvalue of matrix A, we need to determine the null space of the matrix A - λI, where λ is the eigenvalue and I is the identity matrix.
Given a matrix A and its eigenvalues, we can find the eigenvectors associated with each eigenvalue by solving the equation (A - λI)v = 0, where λ is an eigenvalue and v is an eigenvector.
To find the basis for the eigenspace, we need to determine the null space of the matrix A - λI. The null space contains all the vectors v that satisfy the equation (A - λI)v = 0. These vectors form a subspace called the eigenspace corresponding to the eigenvalue λ.
To find a basis for the eigenspace, we can perform Gaussian elimination on the augmented matrix [A - λI | 0] and obtain the reduced row-echelon form. The columns corresponding to the free variables in the reduced row-echelon form will give us the basis vectors for the eigenspace.
For each listed eigenvalue, we repeat this process to find the basis vectors for the corresponding eigenspace. The number of basis vectors will depend on the dimension of the eigenspace, which is determined by the number of free variables in the reduced row-echelon form.
By finding a basis for each eigenspace, we can fully characterize the eigenvectors associated with the given eigenvalues of matrix A.
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Evan and Peter have a radio show that has 2 parts. They need 4 fewer than 11 songs in the first part. In the second part, they need 5 fewer than 3 times the number of songs in the first part. Write an expression for the number of songs they need for their show. A.
(11−4)+3×11−4−5 B. (11−4)+3×(11−4)−5 C. (11−4)+3−4×11−5 D. (11−4)+3−5×(11−4)
Part B How many songs do they need for their show? A. 39 songs B. 31 songs C. 25 songs D. 23 songs.
Answer: they need 28 songs for their show, which corresponds to option D.
Step-by-step explanation:
The expression for the number of songs they need for their show is (11-4) + 3×(11-4) - 5, which corresponds to option B.
To find how many songs they need for their show, we can evaluate the expression:
(11-4) + 3×(11-4) - 5 = 7 + 3×7 - 5 = 7 + 21 - 5 = 28.
We consider the non-homogeneous problem y" + 2y + 5y = 20 cos(z) First we consider the homogeneous problem y" + 2y + 5y = 0: 1) the auxiliary equation is ar² + br + c = <=0. 2) The roots of the auxiliary equation are (enter answers as a comma separated list). (enter answers as a comma separated list). Using these we obtain 3) A fundamental set of solutions is the the complementary solution y = C131 +029/2 for arbitrary constants c₁ and c₂. Next we seek a particular solution y, of the non-homogeneous problem y" + 2y + 5y = 20 cos(z) using the method of undetermined coefficients (See the link below for a help sheet) 4) Apply the method of undetermined coefficients to find yp We then find the general solution as a sum of the complementary solution yc = C131 C232 and a particular solution: y=ye+Up. Finally you are asked to use the general solution to solve an IVP. 5) Given the initial conditions y(0) 5 and y'(0) 5 find the unique solution to the IVP y =
The unique solution to the IVP is:
[tex]y = \frac{35}{6} e^{-t} cos(2t) + \frac{35}{6} e^{-t} sin(2t) - 20[/tex]
How to solve Non - Homogenous Equations?We are given the non-homogeneous problem as:
y" + 2y + 5y = 20 cos(z)
The auxiliary equation is ar² + br + c = 0.
The coefficients for our equation are: a = 1, b = 2, c = 5.
Solving the auxiliary equation, we find the roots:
r = (-b ± √(b² - 4ac)) / (2a)
= (-2 ± √(2² - 4(1)(5))) / (2(1))
= (-2 ± √(-16)) / 2
= (-2 ± 4i) / 2
= -1 ± 2i
The roots of the auxiliary equation are -1 + 2i and -1 - 2i.
A fundamental set of solutions for the homogeneous problem is given by:
y = C₁[tex]e^{-t}[/tex]cos(2t) + C₂[tex]e^{-t}[/tex]sin(2t)
Here, C₁ and C₂ are arbitrary constants.
To find a particular solution ([tex]y_{p}[/tex]) using the method of undetermined coefficients, we assume the form:
y_p = A cos(z) + B sin(z)
where A and B are coefficients to be determined.
Differentiating y_p twice:
y_p" = -A cos(z) - B sin(z)
Substituting y_p and its derivatives into the non-homogeneous equation:
(-A cos(z) - B sin(z)) + 2(A cos(z) + B sin(z)) + 5(A cos(z) + B sin(z)) = 20 cos(z)
Equating the coefficients of cos(z) and sin(z) separately:
-A + 2A + 5A = 0 (coefficients of cos(z))
-B + 2B + 5B = 20 (coefficients of sin(z))
Solving these equations, we find A = -20/6 and B = -10/6.
Therefore, the particular solution is [tex]y_{p}[/tex] = (-20/6)cos(z) - (10/6)sin(z).
The general solution is the sum of the complementary solution (yc) and the particular solution ([tex]y_{p}[/tex]):
y = [tex]y_{c}[/tex] + [tex]y_{p}[/tex]
= C₁[tex]e^{-t}[/tex]cos(2t) + C₂[tex]e^{-t}[/tex]sin(2t) - (20/6)cos(z) - (10/6)sin(z)
To solve the initial value problem (IVP) with the given initial conditions y(0) = 5 and y'(0) = 5, we substitute the initial values into the general solution and solve for the constants C₁ and C₂.
At t = 0:
5 = C₁cos(0) + C₂sin(0) - (20/6)cos(0) - (10/6)sin(0)
5 = C₁ - (20/6)
At t = 0:
5 = -C₁sin(0) + C₂cos(0) + (20/6)sin(0) - (10/6)cos(0)
5 = C₂ - (10/6)
Solving these equations, we find C₁ = 35/6 and C₂ = 35/6.
Therefore, the unique solution to the IVP is:
[tex]y = \frac{35}{6} e^{-t} cos(2t) + \frac{35}{6} e^{-t} sin(2t) - 20[/tex]
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The unique solution to the initial value problem is:
y(z) = 6e^(-z)cos(2z) + 5e^(-z)sin(2z) - cos(z)
To solve the given non-homogeneous problem y" + 2y + 5y = 20cos(z), we can follow the steps outlined:
Homogeneous Problem:
The auxiliary equation for the homogeneous problem y" + 2y + 5y = 0 is:
r² + 2r + 5 = 0
Solving this quadratic equation, we find the roots as complex numbers:
r = -1 + 2i and r = -1 - 2i
Fundamental Set of Solutions:
A fundamental set of solutions for the homogeneous problem is given by:
y_c(z) = C₁e^(-z)cos(2z) + C₂e^(-z)sin(2z), where C₁ and C₂ are arbitrary constants.
Particular Solution:
To find the particular solution, we use the method of undetermined coefficients. Since the right-hand side of the non-homogeneous equation is 20cos(z), we can assume a particular solution of the form:
y_p(z) = Acos(z) + Bsin(z)
Differentiating twice, we find:
y_p''(z) = -Acos(z) - Bsin(z)
Substituting these derivatives into the non-homogeneous equation, we get:
(-Acos(z) - Bsin(z)) + 2(Acos(z) + Bsin(z)) + 5(Acos(z) + Bsin(z)) = 20cos(z)
Simplifying and comparing coefficients of cos(z) and sin(z), we obtain:
-4A + 8B + 20A = 20
8A + 4B + 20B = 0
Solving these equations, we find A = -1 and B = 0.
Therefore, the particular solution is:
y_p(z) = -cos(z)
The general solution is the sum of the complementary solution and the particular solution:
y(z) = y_c(z) + y_p(z)
= C₁e^(-z)cos(2z) + C₂e^(-z)sin(2z) - cos(z)
Initial Value Problem:
To solve the initial value problem with y(0) = 5 and y'(0) = 5, we substitute these values into the general solution and solve for the arbitrary constants.
Given y(0) = 5:
5 = C₁cos(0) + C₂sin(0) - cos(0)
5 = C₁ - 1
Given y'(0) = 5:
5 = -C₁sin(0) + C₂cos(0) + sin(0)
5 = C₂
Therefore, C₁ = 6 and C₂ = 5.
The unique solution to the initial value problem is:
y(z) = 6e^(-z)cos(2z) + 5e^(-z)sin(2z) - cos(z).
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If
Au = -1 4
2 -1 -1 and Av = 3
then
A(3ũ – 3v) =
The value of A(3ũ – 3v), where Au = [-1, 4] and Av = [3] is A(3ũ – 3v) = [6, 18].
The value of A(3ũ – 3v), where Au = [-1, 4] and Av = [3], we first need to determine the matrix A.
That Au = [-1, 4] and Av = [3], we can set up the following system of equations:
A[u₁, u₂] = [-1, 4]
A[v₁, v₂] = [3]
Expanding the system of equations, we have:
A[u₁, u₂] = [-1, 4]
A[3, 0] = [3]
This implies that the second column of A is [0, 4], and the first column of A is [-1, 3].
Therefore, the matrix A can be written as:
A = [ -1, 0 ]
[ 3, 4 ]
Now, we can calculate A(3ũ – 3v):
A(3ũ – 3v) = A(3[u₁, u₂] – 3[v₁, v₂])
= A[3u₁ - 3v₁, 3u₂ - 3v₂]
Substituting the values of u₁, u₂, v₁, and v₂, we have:
A[3(-1) - 3(3), 3(4) - 3(0)]
Simplifying:
A[-6, 12]
To calculate the product, we multiply each element of the matrix A by the corresponding element of the vector [-6, 12]:
A[-6, 12] = [-6*(-1) + 012, 3(-6) + 4*12]
= [6, 18]
Therefore, A(3ũ – 3v) = [6, 18].
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Find the coordinate vector of w relative to the basis S = R². Let u₁ (w) s = = (2, -3), u2 = (3,5), w = = (1,1). (?, ?) (u₁, u₂) for
The coordinate vector of w relative to the basis S = {u₁, u₂} is (a, b) = (1/19, 5/19).
To find the coordinate vector of w relative to the basis S = {u₁, u₂}, we need to express w as a linear combination of u₁ and u₂.
Given:
u₁ = (2, -3)
u₂ = (3, 5)
w = (1, 1)
We need to find the coefficients a and b such that w = au₁ + bu₂.
Setting up the equation:
(1, 1) = a*(2, -3) + b*(3, 5)
Expanding the equation:
(1, 1) = (2a + 3b, -3a + 5b)
Equating the corresponding components:
2a + 3b = 1
-3a + 5b = 1
Solving the system of equations:
Multiplying the first equation by 5 and the second equation by 2, we get:
10a + 15b = 5
-6a + 10b = 2
Adding the two equations:
10a + 15b + (-6a + 10b) = 5 + 2
4a + 25b = 7
Now, we can solve the system of equations:
4a + 25b = 7
We can use any method to solve this system, such as substitution or elimination. For simplicity, let's solve it using substitution:
From the first equation, we can express a in terms of b:
a = (7 - 25b)/4
Substituting this value of a into the second equation:
-3(7 - 25b)/4 + 5b = 1
Simplifying and solving for b:
-21 + 75b + 20b = 4
95b = 25
b = 25/95 = 5/19
Substituting this value of b back into the equation for a:
a = (7 - 25(5/19))/4 = (133 - 125)/76 = 8/76 = 1/19
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Use the present value formula to determine the amount to be invested now, or the present value needed.
The desired accumulated amount is $150,000 after 2 years invested in an account with 6% interest compounded quarterly.
A. The amount to be invested now, or the present value needed, to accumulate $150,000 after 2 years with a 6% interest compounded quarterly is approximately $132,823.87.
B. To determine the present value needed to accumulate a desired amount in the future, we can use the present value formula in compound interest calculations.
The present value formula is given by:
PV = FV / (1 + r/n)^(n*t)
Where PV is the present value, FV is the future value or desired accumulated amount, r is the interest rate (in decimal form), n is the number of compounding periods per year, and t is the number of years.
In this case, the desired accumulated amount (FV) is $150,000, the interest rate (r) is 6% or 0.06, the compounding is quarterly (n = 4), and the investment period (t) is 2 years.
Substituting these values into the formula, we have:
PV = 150,000 / (1 + 0.06/4)^(4*2)
Simplifying the expression inside the parentheses:
PV = 150,000 / (1 + 0.015)^(8)
Calculating the exponent:
PV = 150,000 / (1.015)^(8)
Evaluating (1.015)^(8):
PV = 150,000 / 1.126825
Finally, calculate the present value:
PV ≈ $132,823.87
Therefore, approximately $132,823.87 needs to be invested now (present value) to accumulate $150,000 after 2 years with a 6% interest compounded quarterly.
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help me pls!! (screenshot)
Answer: f(-6) = 44
Step-by-step explanation:
You replace every x with -6
2(-6) squared + 5(-6) - -6/3
36 x 2 -30 + 2
72 - 30 + 2
42 + 2
44
If we use the limit comparison test to determine, then the series Σ 1 n=17+8nln(n) 1 converges 2 limit comparison test is inconclusive, one must use another test. 3 diverges st neither converges nor diverges
The series [tex]$\displaystyle \sum _{ n=17}^{\infty }\dfrac{ 8n\ln( n)}{ n+1}$[/tex] cannot be determined by the limit comparison test and requires another test for convergence.
The limit comparison test is inconclusive in this case. The limit comparison test is typically used to determine the convergence or divergence of a series by comparing it to a known series. However, in this case, it is not possible to find a known series that can be used for comparison. The series [tex]$\displaystyle \sum _{ n=17}^{\infty }\dfrac{ 8n\ln( n)}{ n+1}$[/tex] does not have a clear pattern or a simple known series to compare it with. Therefore, the limit comparison test cannot provide a definitive conclusion.
To determine the convergence or divergence of the series [tex]$\displaystyle \sum _{ n=17}^{\infty }\dfrac{ 8n\ln( n)}{ n+1}$[/tex], one must employ another convergence test. There are several convergence tests available, such as the integral test, ratio test, or root test, which can be applied to this series to determine its convergence or divergence. It is necessary to explore alternative methods to establish the convergence or divergence of this series since the limit comparison test does not yield a conclusive result.
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D² = ( ) x + (0) Find the general solution of Dx= 2t D² = (1 1)² is A(1) - Ge²¹ (1) + 0₂ (1). = C2 You may use that the general solution of D
The general solution of the given differential equation Dx = 2t, with D² = (1 1)², is A(1) - Ge²¹(1) + 0₂(1) = C2.
To find the general solution of the differential equation Dx = 2t, we start by integrating both sides of the equation with respect to x. This gives us the antiderivative of Dx on the left-hand side and the antiderivative of 2t on the right-hand side. Integrating 2t with respect to x yields t² + C₁, where C₁ is the constant of integration.
Next, we apply the operator D² = (1 1)² to the general solution we obtained. This operator squares the derivative and produces a new expression. In this case, (1 1)² simplifies to (2 2).
Now we have D²(t² + C₁) = (2 2)(t² + C₁). Expanding this expression gives us D²(t²) + D²(C₁) = 2t² + 2C₁.
Since D²(t²) = 0 (the second derivative of t² is zero), we can simplify the equation to D²(C₁) = 2t² + 2C₁.
At this point, we introduce the solution A(1) - Ge²¹(1) + 0₂(1) = C₂, where A, G, and C₂ are constants. This is the general solution to the differential equation Dx = 2t, with D² = (1 1)².
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Determine whether the following matrices are in echelon form, reduced echelon form or not in echelon form.
a. Choose
-10 0 1
0 -8 0
b.
Choose
1 0 1
0 1 0
0 0 0
c. Choose
1 0 0 -5
0 1 0 -2
0 0 0 0 d. Choose
1 0 0 4
0 0 0 0
0 1 0 -7
Note: In order to get credit for this problem all answers must be correct.
Problem 14. (a) Perform the indicated row operations on the matrix A successively in the order they are given until a matrix in row echelon form is produced.
A = 3 -9 -3
5 -14 -3
Apply (1/3)R1 → R₁ to A.
Apply R₂-5R1→ R₂ to the previous result.
(b) Solve the system
x=
J 3x1-9x2 = do do
The solution to echelon form matrix of the system is x = (1, -1, -35/3, -14/3, 1)
(a) Let's analyze each matrix to determine if it is in echelon form, reduced echelon form, or not in echelon form:
a. A = | 10 0 10 -8 0 |
| 0 0 0 0 0 |
This matrix is not in echelon form because there are non-zero elements below the leading 1s in the first row.
b. B = | 1 0 10 1 0 |
| 0 0 0 0 0 |
This matrix is in echelon form because all non-zero rows are above any rows of all zeros. However, it is not in reduced echelon form because the leading 1s do not have zeros above and below them.
c. C = | 1 0 0 -50 |
| 1 0 -20 0 |
| 0 0 0 0 |
This matrix is not in echelon form because there are non-zero elements below the leading 1s in the first and second rows.
d. D = | 1 0 0 40 |
| 0 1 0 -7 |
| 0 0 0 0 |
This matrix is in reduced echelon form because it satisfies the following conditions:
All non-zero rows are above any rows of all zeros.
The leading entry in each non-zero row is 1.
The leading 1s are the only non-zero entry in their respective columns.
(b) The system of equations can be written as follows:
3x1 - 9x2 = 0
To solve this system, we can use row operations on the augmented matrix [A | B] until it is in reduced echelon form:
Multiply the first row by (1/3) to make the leading coefficient 1:
R1' = (1/3)R1 = (1/3) * (3 -9 -35 -14 -3) = (1 -3 -35/3 -14/3 -1)
Subtract 5 times the first row from the second row:
R2' = R2 - 5R1 = (0 0 0 0 0) - 5 * (1 -3 -35/3 -14/3 -1) = (-5 15 35/3 28/3 5)
The resulting matrix [A' | B'] in reduced echelon form is:
A' = (1 -3 -35/3 -14/3 -1)
B' = (-5 15 35/3 28/3 5)
From the reduced echelon form, we can obtain the solution to the system of equations:
x1 = 1
x2 = -1
x3 = -35/3
x4 = -14/3
x5 = 1
Therefore, the solution to the system is x = (1, -1, -35/3, -14/3, 1).
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Solve the given problem related to population growth. A city had a population of 23,900 in 2007 and a population of 25,300 in 2012. (a) Find the exponential growth function for the city. Use t=0 to represent 2007 . (Round k to five decimal places.) N(t)= (b) Use the growth function to predict the population of the city in 2022. Round to the nearest hundred.
(a) Here the population growth is exponential and it is given that the population in the year 2007 was 23,900 and population in the year 2012 was 25,300.
The function to predict the population is of the form
N(t) = N0 x (1 + r)t
where,
N0 = initial populationt
= number of yearsr
= growth rate
N(t) = population after t years
From the given data, we can find the growth rate using the formula:
r = (ln P1 - ln P0) / (t1 - t0)
r = (ln 25,300 - ln 23,900) / (2012 - 2007)
r = 0.0237
Then, the exponential growth function is given by:
N(t) = N0 x (1 + r)tN(t)
= 23,900 x (1 + 0.0237)tN(t)
= 23,900 x 1.0237t
(b) Predict the population of the city in 2022Using the growth function:
N(t) = 23,900 x 1.0237t
If t = 2022 - 2007
= 15 yearsN(15)
= 23,900 x 1.023715
≈ 30,200
Hence, the population of the city in 2022 is approximately 30,200.
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K- 3n+2/n+3 make "n" the Subject
The expression "n" as the subject is given by:
n = (2 - 3K)/(K - 3)
To make "n" the subject in the expression K = 3n + 2/n + 3, we can follow these steps:
Multiply both sides of the equation by (n + 3) to eliminate the fraction:
K(n + 3) = 3n + 2
Distribute K to both terms on the left side:
Kn + 3K = 3n + 2
Move the terms involving "n" to one side of the equation by subtracting 3n from both sides:
Kn - 3n + 3K = 2
Factor out "n" on the left side:
n(K - 3) + 3K = 2
Subtract 3K from both sides:
n(K - 3) = 2 - 3K
Divide both sides by (K - 3) to isolate "n":
n = (2 - 3K)/(K - 3)
Therefore, the expression "n" as the subject is given by:
n = (2 - 3K)/(K - 3)
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Suppose $30,000 is deposited into an account paying 4.5% interest, compounded continuously. How much money is in the account after 8 years if no withdrawals or additional deposits are made?
There is approximately $41,916 in the account after 8 years if no withdrawals or additional deposits are made.
To calculate the amount of money in the account after 8 years with continuous compounding, we can use the formula [tex]A = P * e^{(rt)}[/tex], where A is the final amount, P is the principal amount (initial deposit), e is Euler's number (approximately 2.71828), r is the interest rate, and t is the time in years.
In this case, the principal amount is $30,000 and the interest rate is 4.5% (or 0.045 in decimal form).
We need to convert the interest rate to a decimal by dividing it by 100.
Therefore, r = 0.045.
Plugging these values into the formula, we get[tex]A = 30000 * e^{(0.045 * 8)}[/tex]
Calculating the exponential part, we have
[tex]e^{(0.045 * 8)} \approx 1.3972[/tex].
Multiplying this value by the principal amount, we get A ≈ 30000 * 1.3972.
Evaluating this expression, we find that the amount of money in the account after 8 years with continuous compounding is approximately $41,916.
Therefore, the answer to the question is that there is approximately $41,916 in the account after 8 years if no withdrawals or additional deposits are made.
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Paris has a utility function over berries (denoted by B ) and chocolate (denoted by C) as follows: U(B, C) = 2ln(B) + 4ln(C) The price of berries and chocolate is PB and pc, respectively. Paris's income is m. 1. What preferences does this utility function represent? 2. Find the MRSBC as a function of B and C assuming B is on the x-axis. 3. Find the optimal bundle B and C as a function of income and prices using the tangency condition. 4. What is the fraction of total expenditure spent on berries and chocolate out of total income, respectively? 5. Now suppose Paris has an income of $600. The price of a container of berries is $10 and the price of a chocolate bar is $10. Find the numerical answers for the optimal bundle, by plugging the numbers into the solution you found in Q3.3.
5. The numerical answers for the optimal bundle of B and C is (75, 37.5).
1 Preferences: The utility function U(B, C) = 2ln(B) + 4ln(C) represents a case of perfect substitutes.
2. MRSBC as a function of B and C: The marginal rate of substitution (MRS) of B for C can be calculated as follows:
MRSBC = ΔC / ΔB = MU_B / MU_C = 2B / 4C = B / 2C
3. Optimal bundle of B and C: To find the optimal bundle of B and C, we use the tangency condition. According to this condition:
MRSBC = PB / PC
This implies that C / B = PB / (2PC)
The budget constraint of the consumer is given by:
m = PB * B + PC * C
The budget line equation can be expressed as:
C = (m / PC) - (PB / PC) * B
But we also have C / B = PB / (2PC)
By substituting the expression for C from the budget line, we can solve for B:
(m / PC) - (PB / PC) * B = (PB / (2PC)) * B
B = (m / (PC + 2PB))
By substituting B in terms of C in the budget constraint, we get:
C = (m / PC) - (PB / PC) * [(m / (PC + 2PB)) / (PB / (2PC))]
C = (m / PC) - (m / (PC + 2PB))
4. Fraction of total expenditure spent on berries and chocolate: Total expenditure is given by:
m = PB * B + PC * C
Dividing both sides by m, we get:
(PB / m) * B + (PC / m) * C = 1
Since the optimal bundle is (B, C), the fraction of total expenditure spent on berries and chocolate is given by the respective coefficients of the bundle:
B / m = (PB / m) * B / (PB * B + PC * C)
C / m = (PC / m) * C / (PB * B + PC * C)
5. Numerical answer for the optimal bundle:
Given:
Income m = $600
Price of a container of berries PB = $10
Price of a chocolate bar PC = $10
Substituting these values into the optimal bundle equation derived in step 3, we get:
B = (600 / (10 + 2 * 10)) = 75 units
C = (1/2) * B = (1/2) * 75 = 37.5 units
Therefore, the optimal bundle of B and C is (75, 37.5).
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Simplify the quantity negative 7 times a to the 3rd power times b to the negative 3 power end quantity divided by the quantity 21 times a times b end quantity.
The simplified form of the expression [tex](-7a^3b^{-3})[/tex] / (21ab) is [tex](-a^2) / (3b^3),[/tex] removing negative exponents and canceling out common factors.
To simplify the given expression, let's break it down step by step.
The expression is:
[tex](-7a^3b^{-3}) / (21ab)[/tex]
First, we can simplify the numerator by applying the exponent rules.
The negative exponent in the numerator can be rewritten as a positive exponent in the denominator:
[tex](-7a^3) / (21ab^3)[/tex]
Next, we can simplify the fraction by canceling out common factors in the numerator and denominator. In this case, we can cancel out the common factor of 7:
[tex](-a^3) / (3ab^3)[/tex]
Now, we can simplify the remaining terms by canceling out the common factor of 'a':
[tex](-a^2) / (3b^3)[/tex]
Finally, we have simplified the expression to [tex](-a^2) / (3b^3)[/tex].
In this simplified form, the expression no longer contains negative exponents or common factors in the numerator and denominator.
To summarize, the simplified form of the expression [tex](-7a^3b^{-3}) / (21ab)[/tex] is [tex](-a^2) / (3b^3).[/tex]
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Rio guessed she would score a 90 on her math test. She earned an 86 on her math test. What is the percent error?
Answer:
4.44%
Step-by-step explanation:
%Error = [tex]\frac{E-T}{T}[/tex] x 100
E = experiment
T = Theoretical
E = 86
T = 90
What is the percent error?
We Take
[tex]\frac{86-90}{90}[/tex] x 100 ≈ 4.44%
So, the percent error is about 4.44%
3. Calculate the Fourier series equation for the equation
0 -2
f(x) = 1 -1
0 1< t <2
The Fourier series equation for the given function f(x) = 1 on the interval 1 < t < 2 is simply f(x) = 0.
To calculate the Fourier series equation for the given function f(x) = 1 on the interval 1 < t < 2, we can follow these steps:
Step 1: Determine the period:
The given interval is 1 < t < 2, which has a length of 1 unit. Since the function is not periodic within this interval, we need to extend it periodically.
Step 2: Extend the function periodically:
We can extend the function f(x) = 1 to be periodic by repeating it outside the interval 1 < t < 2. Let's extend it to the interval -∞ < t < ∞, such that f(x) remains constant at 1 for all values of t.
Step 3: Determine the Fourier coefficients:
To find the Fourier coefficients, we need to calculate the integral of the function multiplied by the corresponding trigonometric functions.
The Fourier coefficient a0 is given by:
a0 = (1/T) * ∫[T] f(t) dt,
where T is the period. Since we have extended the function to be periodic over all t, the period T is infinite.
The integral becomes:
a0 = (1/∞) * ∫[-∞ to ∞] 1 dt = 1/∞ = 0.
The Fourier coefficients an and bn are given by:
an = (2/T) * ∫[T] f(t) * cos(nωt) dt,
bn = (2/T) * ∫[T] f(t) * sin(nωt) dt,
where ω = 2π/T.
Since T is infinite, the integrals become:
an = (2/∞) * ∫[-∞ to ∞] 1 * cos(nωt) dt = 0,
bn = (2/∞) * ∫[-∞ to ∞] 1 * sin(nωt) dt = 0.
Step 4: Write the Fourier series equation:
The Fourier series equation for the given function is:
f(x) = a0/2 + ∑[n=1 to ∞] (an * cos(nωt) + bn * sin(nωt)).
Substituting the Fourier coefficients we calculated, we have:
f(x) = 0/2 + ∑[n=1 to ∞] (0 * cos(nωt) + 0 * sin(nωt)).
Simplifying, we get:
f(x) = 0.
Therefore, the Fourier series equation for the given function f(x) = 1 on the interval 1 < t < 2 is simply f(x) = 0.
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Let A = 3 2 3-4-5 3 1 a) Find a basis for the row space of A. b) Find a basis for the null space of A. c) Find rank(A). d) Find nullity (A).
A basis for the row space of A is {[1, 0, -1, 4, 5], [0, 1, 2, -2, -2]}. A basis for the null space of A is {[-1, -2, 1, 0, 0], [4, 2, 0, 1, 0], [-5, 2, 0, 0, 1]}. The rank of A is 2. The nullity of A is 3.
a) To find a basis for the row space of A, we row-reduce the matrix A to its row-echelon form.
Row reducing A, we have:
R = 1 0 -1 4 5
0 1 2 -2 -2
0 0 0 0 0
The non-zero rows in the row-echelon form R correspond to the non-zero rows in A. Therefore, a basis for the row space of A is given by the non-zero rows of R: {[1, 0, -1, 4, 5], [0, 1, 2, -2, -2]}
b) To find a basis for the null space of A, we solve the homogeneous equation Ax = 0.
Setting up the augmented matrix [A | 0] and row reducing, we have:
R = 1 0 -1 4 5
0 1 2 -2 -2
0 0 0 0 0
The parameters corresponding to the free variables in the row-echelon form R are x3 and x5. We can express the dependent variables x1, x2, and x4 in terms of these free variables:
x1 = -x3 + 4x4 - 5x5
x2 = -2x3 + 2x4 + 2x5
x4 = x3
x5 = x5
Therefore, a basis for the null space of A is given by the vector:
{[-1, -2, 1, 0, 0], [4, 2, 0, 1, 0], [-5, 2, 0, 0, 1]}
c) The rank of A is the number of linearly independent rows in the row-echelon form R. In this case, R has two non-zero rows, so the rank of A is 2.
d) The nullity of A is the dimension of the null space, which is equal to the number of free variables in the row-echelon form R. In this case, R has three columns corresponding to the free variables, so the nullity of A is 3.
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A statistics student is interested in the relationship between the size of a pizza (the diameter measured in inches) and its price. He collects a random sample of pizzas from several local restaurants. He finds a linear model to give the relationship between the size of the pizza and the price. The equation of the line is ŷ = –8.1 + 1.91x, where ŷ is the price and x is the diameter. The residual plot is shown.
The correct statement regarding the residuals is given as follows:
Yes, the residuals are relatively small.
What are residuals?For a data-set, the definition of a residual is that it is the difference of the actual output value by the predicted output value, that is:
Residual = Observed - Predicted.
Hence, on the graph, the residuals are given by the vertical distance between each point on the line.
The points are close to the line in this problem, meaning that the residuals are small and the model is a good fit.
More can be learned about residuals at brainly.com/question/1447173
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Geno read 126 pages in 3 hours. He read the same number of pages each hour for the first 2 hours. Geno read 1. 5 times as many pages during the third hour as he did during the first hour.
Let's assume that Geno read x pages each hour for the first 2 hours. Geno read 36 pages each hour for the first two hours and 1.5 times as many, during the third hour.
During the first hour, Geno read x pages. During the second hour, Geno read x pages again. So, in the first two hours, Geno read a total of 2x pages. According to the given information, Geno read 1.5 times as many pages during the third hour as he did during the first hour. Therefore, during the third hour, he read 1.5x pages.
In total, Geno read 2x + 1.5x = 3.5x pages in 3 hours.
We also know that Geno read 126 pages in total.
Therefore, we can set up the equation: 3.5x = 126.
Solving this equation, we find x = 36.
So, Geno read 36 pages each hour for the first two hours and 1.5 times as many, which is 54 pages, during the third hour.
Learn more about hours here
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