If g(6) = 3 - 5(6), what is g(x)?

To prove F(c)(A + B) → (Vx) A (3.c)B, we need to show that if the union of sets A and B is finite, then there exists an element x in A such that for all elements y in B, (x, y) is in the relation C.

Assume F(c)(A + B) is true. Then for any (x, y) in C, x belongs to A + B, which means x belongs to either A or B. If x belongs to A, then we have found an element x in A such that for all elements y in B, (x, y) is in C, and we are done. If x belongs to B, then we need to find another element in A such that the condition holds.

Since A and B are finite, their union A + B is also finite. Let n be the size of A + B. Then there are n distinct elements in A + B, say a1, a2, ..., an. Since there are more elements in A than in B (or equal if they have the same size), there must be at least one element of A among a1, a2, ..., an. Call this element x.

Now, consider any element y in B. Since x belongs to A + B and y belongs to B, their sum x + y belongs to A + B as well. But we know that x + y cannot be equal to x, since y is not in A. Therefore, x + y must be equal to one of the remaining n-1 elements of A + B, say ai. But then ai - x = y, so (x, y) is in C.

Therefore, we have shown that F(c)(A + B) → (Vx) A (3.c)B is true.

For the second part of the question, we need to show that for any set A in N, there exists a set B in N such that A is a subset of B and B is infinite.

Let B be the set of all natural numbers greater than the maximum element in A. Then A is clearly a subset of B, and B is infinite since it contains all natural numbers greater than a certain number.

Therefore, we have shown that for any set A in N, there exists a set B in N such that A is a subset of B and B is infinite.

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(a) When the ball leaves the child's hand, x = 0, so we can substitute this into the equation:
y = -1/16(0)^2 + 2(0) + 5
y = 5
Therefore, the ball is 5 feet high when it leaves the child's hand.

(b) To find the maximum height of the ball, we need to determine the vertex of the parabola. The x-coordinate of the vertex is given by:
x = -b/2a
where a = -1/16 and b = 2. Substituting these values:
x = -2/(2(-1/16))
x = 16
To find the y-coordinate, we substitute x = 16 into the equation:
y = -1/16(16)^2 + 2(16) + 5
y = 21
Therefore, the maximum height of the ball is 21 feet.

(c) To find how far from the child the ball strikes the ground, we need to determine the value of x when y = 0. Substituting y = 0 into the equation:
0 = -1/16x^2 + 2x + 5
Multiplying both sides by -16 to eliminate the fraction:
0 = x^2 - 32x - 80
We can solve for x using the quadratic formula:
x = (32 ± sqrt(32^2 - 4(1)(-80))) / 2(1)
x = (32 ± sqrt(1472)) / 2
x = 16 ± 8sqrt(2)
Since the ball cannot land behind the child, we take the positive value:
x = 16 + 8sqrt(2)
Therefore, the ball strikes the ground approximately 29.1 feet from the child.

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Answer:

B: 4.5

Step-by-step explanation:

To evaluate the expressions for z = 1 + i, we'll consider each part:

(a) zz (the product of z and its complex conjugate):
First, find the complex conjugate of z, which is the same as z but with the imaginary part negated: z* = 1 - i. Now, multiply z and z*:
z * z* = (1 + i)(1 - i) = 1 - i + i - i^2 = 1 - i^2 = 1 - (-1) = 1 + 1 = 2.

(b) Re(z) + Im(z) Z^2 + 2:
Re(z) is the real part of z, which is 1. Im(z) is the imaginary part of z, which is 1. Now, square z:
Z^2 = (1 + i)^2 = 1 + 2i + i^2 = 1 + 2i - 1 = 2i.
Finally, calculate Re(z) + Im(z) Z^2 + 2:
1 + 1(2i) + 2 = 1 + 2i + 2 = 3 + 2i.

So, the expressions evaluated are (a) zz = 2 and (b) Re(z) + Im(z) Z^2 + 2 = 3 + 2i.

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The answer is not provided in the options given. The closest option is 0.172, but the correct answer is 0.155 (rounded to three decimal places).

To calculate the standard deviation of X, we first need to find the mean or expected value of X. We can do this by integrating the given pdf over the range 0.90 to 1.80:

E(X) = ∫[0.90,1.80] x*f(x) dx

= ∫[0.90,1.80] x*(265/652)*x^3 dx

= (265/652) * ∫[0.90,1.80] x^4 dx

= (265/652) * [x^5/5] from x=0.90 to x=1.80

≈ 1.315

Next, we can calculate the variance of X using the formula:

Var(X) = E(X^2) - [E(X)]^2

To find E(X^2), we integrate the pdf squared over the same range:

E(X^2) = ∫[0.90,1.80] x^2*f(x) dx

= ∫[0.90,1.80] x^2*(265/652)*x^3 dx

= (265/652) * ∫[0.90,1.80] x^5 dx

= (265/652) * [x^6/6] from x=0.90 to x=1.80

≈ 1.464

Var(X) = E(X^2) - [E(X)]^2

≈ 1.464 - 1.315^2

≈ 0.024

Finally, we take the square root of the variance to obtain the standard deviation:

sigma = sqrt(Var(X))

≈ sqrt(0.024)

≈ 0.155

Therefore, the answer is not provided in the options given. The closest option is 0.172, but the correct answer is 0.155 (rounded to three decimal places).

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8% for 24 months has the lower finance charge of $296, compared to option b) with a finance charge of $363.50.

To compare the two loans, we need to calculate the finance charge for each option.

For option a) at 8% APR for 24 months, we can use the following formula:

Finance charge = (loan amount x interest rate x time) / 12

Finance charge = (1850 x 0.08 x 24) / 12 = 296

So the finance charge for option a) is $296.

For option b) at 11% APR for 18 months, we can use the same formula:

Finance charge = (loan amount x interest rate x time) / 12

Finance charge = (1850 x 0.11 x 18) / 12 = 363.5

So the finance charge for option b) is $363.50.

Therefore, option a) has the lower finance charge of $296, compared to option b) with a finance charge of $363.50.

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To find the probability that the amount of pollutant in a given sample will exceed 1/2 for the smokestack with the filter installed, you need to determine the distribution of the pollutant levels and then calculate the probability based on that distribution.

To find the probability that the amount of pollutant in a given sample will exceed 1/2 when a filter is installed in the smokestack, we need to use the information provided in the question. However, we do not have any specific information on the distribution of the pollutant levels, so we cannot calculate the exact probability.
Instead, we can make some assumptions based on the purpose of the filter. Filters are typically installed to reduce the amount of pollutants emitted into the air, so it is reasonable to assume that the filter will decrease the amount of pollutant in each sample. Therefore, we can expect the probability of the pollutant level exceeding 1/2 to decrease when a filter is installed.
Without more information, we cannot give an exact probability, but we can say that it is likely lower than the probability without a filter. We would need to know more about the specific characteristics of the filter and the pollutant to make a more accurate estimate.

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Hi, I'm glad to help you with this question. To determine if the normal distribution gives a reasonable approximation using the given data, we need to calculate the mean and standard deviation. Here are the steps:

1. Calculate the mean (average): Add all the data points together and divide by the number of data points.
(71+42+77+84+46+93+94+63+82+88+57+32+79+67+68+83+60+65+58+70) / 20 = 1380 / 20 = 69

Mean = 69

2. Calculate the standard deviation: First, find the difference between each data point and the mean, square the differences, and then find the average of those squared differences. Finally, take the square root of that average.
a. Differences from the mean: (-2, 27, 8, 15, -23, 24, 25, -6, 13, 19, -12, -37, 10, -2, -1, 14, -9, -4, -11, 1)
b. Squared differences: (4, 729, 64, 225, 529, 576, 625, 36, 169, 361, 144, 1369, 100, 4, 1, 196, 81, 16, 121, 1)
c. Average of squared differences: (4520) / 20 = 226
d. Square root of the average: √226 ≈ 15.03

Standard Deviation ≈ 15.03

Now that we have the mean (69) and the standard deviation (15.03), you can compare these values to the distribution line that you draw through the data by eye. If the distribution line follows a bell-shaped curve with the mean at the center and the data points spread around it following the standard deviation, then the normal distribution provides a reasonable approximation for this data set.

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The equations are;

H = 24 + 9x

H = 40 + 5x

In a word problem, there are usually one or more unknown quantities that you need to find. Identify these unknowns and assign them a variable.

We have to know that Tree A was 24 inches tall and Tree B was 40 inches tall. Each year thereafter, Tree A grew by 9 inches per year and Tree B grew by 5 inches per year.

Then for tree A;

H = 24 + 9x

For tree B

H = 40 + 5x

Where x is the number of years that the trees stay.

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The theoretical probability of the spinner not landing on blue would be = 0.625. That is option B.

To calculate the theoretical probability of the given event, the formula that should be used is given as follows:

Probability = possible outcome/sample space

The possible outcome for other colours apart from blue = 5

The sample space = 8

Therefore probability = 5/8 = 0.625

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The solutions to the equation to the nearest tenth are x = 10.1 and x = 2.9.

We have,

-2x² - 13x + 20 = -3x²

Combining like terms

-2x² - 13x + 20 = -3x²

x² - 13x + 20 = 0 (adding 3x² to both sides)

Now we can use the quadratic formula to solve for x:

x = (-b ± √(b² - 4ac)) / 2a

In this case,

a = 1, b = -13, and c = 20.

Substituting these values into the quadratic formula:

x = (-(-13) ± √((-13)² - 4(1)(20))) / 2(1)

x = (13 ± √(169 - 80)) / 2

x = (13 ± √(89)) / 2

So the solutions are:

x = (13 + √(89)) / 2

x ≈ 10.1

and

x = (13 - √(89)) / 2

x ≈ 2.9

Therefore,

The solutions to the equation to the nearest tenth are x ≈ 10.1 and x ≈ 2.9.

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The expected value of a $4 bet on the state's lottery is -$0.84 and the expected value of a $4 bet on the casino game is -1.7¢ (which is equivalent to -$0.017), the state's lottery is a better bet in terms of producing a higher expected value.

a. There are 10 possible choices for each of the three digits, so the total number of different selections is 10 x 10 x 10 = 1000.

b. Since there is only one winning combination out of the 1000 possible selections, the probability of winning is 1/1000.

c. If you win, your net profit would be $500 - $4 = $496.

d. The expected value is the sum of the products of each possible outcome and its probability. In this case, the expected value is (1/1000) x $500 + (999/1000) x (-$4) = -$0.84.

e. Since the expected value of a $4 bet on the state's lottery is -$0.84 and the expected value of a $4 bet on the casino game is -1.7¢ (which is equivalent to -$0.017), the state's lottery is a better bet in terms of producing a higher expected value.

This is because the expected loss for the state's lottery is smaller than the expected loss for the casino game.

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The coupon rate for the semi-annual bond is approximately 11.96%.

Given the information provided, we have the following details:

- Par value: $1,000
- Maturity: 10 years
- Current price: $887
- YTM (Yield to Maturity): 10.9%

To solve for the coupon rate, we can use the bond pricing formula:

Bond Price = (C * (1 - (1 + r/2)^(-2n))) / (r/2) + (Par Value / (1 + r/2)^(2n))

Where:
- Bond Price = $887
- C = Coupon payment per period (which we need to find)
- r = YTM / 100 = 0.109
- n = Maturity in years = 10

Plugging in the given values:

$887 = (C * (1 - (1 + 0.109/2)^(-2*10))) / (0.109/2) + ($1,000 / (1 + 0.109/2)^(2*10))

Now, we can solve for the coupon payment, C:

C = (($887 * 0.109/2) - ($1,000 / (1 + 0.109/2)^(2*10))) / (1 - (1 + 0.109/2)^(-2*10))

C ≈ $59.80

Since this is a semi-annual bond, the annual coupon payment would be:

Annual Coupon Payment = C * 2 = $59.80 * 2 = $119.60

Finally, to find the coupon rate, we can divide the annual coupon payment by the par value:

Coupon Rate = (Annual Coupon Payment / Par Value) * 100 = ($119.60 / $1,000) * 100 = 11.96%

So, the coupon rate for the semi-annual bond is approximately 11.96%.

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The measure of the dilation image P'Q' with a scale factor of 3/4 is given as follows:

P'Q' = 9 units.

A dilation can be defined as a transformation that multiplies the distance between every point in an object and a fixed point, called the center of dilation, by a constant factor called the scale factor.

The scale factor for the dilation in this problem is given as follows:

k = 3/4.

The length of the original segment is of 12 units, hence the length of the dilated segment is given as follows:

P'Q' = 3/4 x 12 = 36/4 = 9 units.

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The value of the given expression is -4, which is integer. Therefore, option B is the correct answer.

The given expression is (p²+q)/(-|p|-q).

Here, p=-6 and q=4.

Substitute p=-6 and q=4 in the given expression we get

((-6)²+4)/(-|-6|-4)

= 40/(-10)

= -4

Therefore, option B is the correct answer.

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The three options with dimensions resulting in a surface area greater than 140 square feet are:

To determine whether the dimensions of a rectangular prism result in a surface area greater than 140 square feet, we can use the formula for the surface area of a rectangular prism:

Surface Area = 2lw + 2lh + 2wh

    where l, w, and h are the length, width, and height of the rectangular prism, respectively.

Option 1: 6 feet by 2 feet by 3 feet

Surface Area = 2(6)(2) + 2(6)(3) + 2(2)(3) = 24 + 36 + 12 = 72 square feet

This option does not have a surface area greater than 140 square feet.

Option 2: 6 feet by 5 feet by 4 feet

Surface Area = 2(6)(5) + 2(6)(4) + 2(5)(4) = 60 + 48 + 40 = 148 square feet

This option has a surface area greater than 140 square feet.

Option 3: 7 feet by 6 feet by 4 feet

Surface Area = 2(7)(6) + 2(7)(4) + 2(6)(4) = 84 + 56 + 48 = 188 square feet

This option has a surface area greater than 140 square feet.

Option 4: 8 feet by 3 feet by 7 feet

Surface Area = 2(8)(3) + 2(8)(7) + 2(3)(7) = 48 + 112 + 42 = 202 square feet

This option has a surface area greater than 140 square feet.

Option 5: 8 feet by 4 feet by 3 feet

Surface Area = 2(8)(4) + 2(8)(3) + 2(4)(3) = 64 + 48 + 24 = 136 square feet

This option does not have a surface area greater than 140 square feet.

Therefore, the three options with dimensions resulting in a surface area greater than 140 square feet are:

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Answer:

1.5 and 4

Step-by-step explanation:

the scale factor is the ratio of corresponding sides, image to original.

1

scale factor = [tex]\frac{DF}{AC}[/tex] = [tex]\frac{21}{14}[/tex] = [tex]\frac{3}{2}[/tex] = 1.5

2

scale factor = [tex]\frac{8}{2}[/tex] = 4

The correlation coefficient of 0.88 reveals that there is an intense linear connection between the two variables.

It should be noted that to ascertain the correlation coefficient, we can utilize the formula:

r = (nΣxy - ΣxΣy) / sqrt[(nΣx^2 - (Σx)^2)(nΣy^2 - (Σy)^2)]

Retaining the same numeric values from before, we can compute:

r = (9(976) - (36)(605)) / sqrt[(9(182) - (36)^2)(9(11681) - (605)^2)] ≈ 0.88

Therefore, the correlation coefficient is fairly close to 0.88.

The correlation coefficient implies a strong positive relationship between hours expended studying and test outcomes. As the quantity of hours committed to studying increases, the test scores will tend to keep up accordingly. A correlation coefficient of 0.88 reveals that there is an intense linear connection between the two variables, and the line of best fit serves as a desirable standard representation of the data.

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Answer:

Step-by-step explanation:

It appears there is an error in the table. The correct table is:

[tex]\begin{array}{|l|c|c|c|c|c|c|c|c|c|c|c|}\cline{1-11}\vphantom{\dfrac12}\textsf{Hours spent studying $(x)$}&0&1&2&4&4&4&6&6&7&8\\\cline{1-11}\vphantom{\dfrac12}\textsf{Test score $(y)$}&35&40&46&65&67&70&82&88&82&95\\\cline{1-11}\end{array}[/tex]

The simplest method to find the linear regression equation and the correlation coefficient for this data set is to use a statistical calculator.

After entering the data into a statistical calculator we get:

The regression line of y on x is y = a + bx.

Therefore, substitute the found values of a and b into the formula to write the linear regression equation for the given data set:

[tex]\boxed{y=34.27+7.79x}[/tex]

The correlation coefficient is the value of r, so r = 0.98 to the nearest hundredth.

The correlation coefficient, r, measures the strength of the linear correlation between two variables. |t is always between +1 and -1.

As r = 0.98, there is a very strong positive correlation.

In context, this suggests that the more hours of studying for a test a student does, the higher their test score.

The probability of getting an odd number or a multiple of 4 is 3/4.

Given that, a twelve-sided die has sides numbered 1 through 12. we need to find the probability of getting a number odd or a multiple of 4.

Probability = favorable outcomes / total number of outcomes

For odd numbers =

Favorable outcomes = 1, 3, 5, 7, 9, 11 = 6

P(odd number) = 6/12 = 1/2

For multiple of 4 =

Favorable outcomes = 4, 8, 12 = 3

P(multiple of 4) = 3/12 = 1/4

P(odd or a multiple of 4) = 1/2 + 1/4 = 3/4

Hence, the probability of getting an odd number or a multiple of 4 is 3/4.

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Answer:

all yes

Step-by-step explanation:

Answer:

option B: x= 18

Step-by-step explanation:

To solve 1/3x - 1 = 5, we can start by adding 1 to both sides of the equation:

1/3x - 1 + 1 = 5 + 1

Simplifying:

1/3x = 6

Multiplying both sides by 3:

3(1/3x) = 3(6)

Simplifying:

x = 18

Therefore, the solution is x = 18, which is option B.

If in a survey, 200 college students were asked whether they live on campus and if they own a car, 55% of college students in the survey don't own a car.

To find the percent of college students who don't own a car, we need to add up the number of students who don't own a car and divide it by the total number of students in the survey. In this case, the total number of students in the survey is 200.

From the table, we can see that there are 88 students who live on campus and don't own a car, and 22 students who don't live on campus and don't own a car. So the total number of students who don't own a car is 88 + 22 = 110.

To find the percentage, we divide the number of students who don't own a car by the total number of students in the survey and then multiply by 100 to get the percentage:

Percentage of students who don't own a car = (110/200) x 100% = 55%

When working with percentages, we need to divide the number we are interested in by the total and then multiply by 100 to get the percentage.

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Means that the derivative of A at a, dA(a), is equal to L. Hence, A is differentiable at a with dA(a) = L.

To prove that the function A is differentiable at a with dA(a) = L, we need to show that:

lim(x→a) [A(x) - A(a) - L(a)(x-a)] / ||x-a|| = 0

We know that A(x) = c + L(x) for all x in RP, where c is a constant and L is a linear transformation from RP to R9.

Then, we have:

A(a) = c + L(a)

L(a)(x-a) = L(x-a) + L(a-a) = L(x-a)

Substituting these into the limit expression, we get:

lim(x→a) [c + L(x) - c - L(a) - L(x-a)] / ||x-a||

= lim(x→a) [L(x) - L(a)] / ||x-a||

Since L is a linear transformation, it is continuous. Therefore, we can write:

lim(x→a) [L(x) - L(a)] / ||x-a|| = L( lim(x→a) [x-a] / ||x-a|| )

But lim(x→a) [x-a] / ||x-a|| = u, a unit vector in the direction of x-a.

Therefore, we have:

lim(x→a) [L(x) - L(a)] / ||x-a|| = Lu

This means that the derivative of A at a, dA(a), is equal to L. Hence, A is differentiable at a with dA(a) = L.

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The probability that a student studied for 5 hours is given as follows:

0.23.

A probability is calculated as the division of the desired number of outcomes by the total number of outcomes in the context of a problem/experiment.

The total number of students in this problem is given as follows:

1 + 3 + 2 + 5 + 9 + 7 + 3 = 30.

Out of those 30 students, 7 studied five hours, hence the probability is given as follows:

p = 7/30 = 0.23.

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Answer:

43.96 cm

Step-by-step explanation:

The circumference of a circle is given by the formula:

C = 2πr

where r is the radius of the circle and π is a mathematical constant approximately equal to 3.14.

Given that the radius of the circle is 7 cm, we can substitute this value into the formula and simplify:

C = 2πr

C = 2 × 3.14 × 7

C = 43.96

Therefore, the circumference of the circle is approximately 43.96 cm.

Answer:

The surface area is 70 yards

Step-by-step explanation:

The formula for surface area is (SA)=2lw+2lh+2hw. Meaning it would be 2 times (8 times 3 + 8 times 1 + 3 times 1) which equals 70 yards.

Answer:

Part A:

Sunday newspaper: Start at $36 and add $4 for each paper delivered.

Saturday newspaper: Start at $12 and add $2.50 for each paper delivered.

Part B:

To calculate Stevic's total earnings after delivering 15 papers on each day:

Earnings from Sunday papers = $36 + ($4 x 15) = $96

Earnings from Saturday papers = $12 + ($2.50 x 15) = $49.50

Total earnings = Earnings from Sunday papers + Earnings from Saturday papers

Total earnings = $96 + $49.50

Total earnings = $145.50

Therefore, Stevic's total earnings after delivering 15 papers on each day is $145.50.

Step-by-step explanation:

There is enough evidence to show that the mean carbon dioxide emission is now lower than it was in 2004.

To test whether the mean carbon dioxide emission is now lower than it was in 2004, we need to conduct a one-sample t-test.

We are given a random sample of carbon dioxide emissions from a recent study. Let's assume that this sample is representative of the population of interest. The null hypothesis is that the true population mean of carbon dioxide emissions is equal to or greater than the mean in 2004 (4.87 metric tons per capita). The alternative hypothesis is that the true population mean is less than the mean in 2004.

We can set up the hypotheses as follows:

H0: μ >= 4.87

Ha: μ < 4.87

where μ is the true population mean of carbon dioxide emissions.

We are given the sample data in a table, but we don't know the population standard deviation, so we will use the sample standard deviation to estimate it. The sample mean is calculated as:

x = (4.28 + 3.94 + 3.27 + 3.81 + 3.43 + 3.09 + 2.52 + 2.98 + 3.23 + 3.36) / 10 = 3.43

The sample standard deviation is calculated as:

s = √(((4.28 -x)² + (3.94 - x)² + ... + (3.36 - x)²) / 9) = 0.659

The sample size is n = 10.

We can calculate the t-statistic as:

t = (x- μ) / (s / √(n)) = (3.43 - 4.87) / (0.659 / √(10)) = -4.26

The degrees of freedom for this test are df = n - 1 = 9. We can use a t-distribution table or a calculator to find the p-value associated with this t-statistic and degrees of freedom.

Using a t-distribution table with df = 9, we find that the p-value for a one-tailed test at the 3% level is less than 0.001. This means that the probability of observing a t-statistic as extreme as -4.26, assuming the null hypothesis is true, is less than 0.001.

Since the p-value is less than the significance level of 0.03, we reject the null hypothesis and conclude that there is enough evidence to show that the mean carbon dioxide emission is now lower than it was in 2004 at the 3% level.

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If sales are expressed as a function of the amount spent on advertising, then the dollar amount at which the rate of change of sales goes from increasing to decreasing is called the point of diminishing returns or diminishers returns to scale.

It sounds like you want to know about the relationship between sales, advertising, and the rate of change. Here's an answer incorporating the terms you've mentioned:

If sales are expressed as a function of the amount spent on advertising, the dollar amount at which the rate of change of sales goes from increasing to decreasing is called the inflection point. If 'x' is that dollar amount, then 'x' represents the advertising budget at which the sales growth rate starts to decline.

This point indicates that increasing the amount spent on advertising beyond this dollar amount will result in a decrease in the rate of change of sales. This is known as the diminishers' scale to return.

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the solution to the given differential equation is: Q(t) = -t^4/25 + t^4/5 ln(t) + C/t

To solve the given differential equation t dQ/dt + Q = t^4 ln(t), we'll first find the integrating factor, solve for Q(t), and then substitute the given terms.

Step 1: Find the integrating factor.
The integrating factor is e^(∫P(t)dt), where P(t) = 1/t in this case. So,

∫(1/t)dt = ln(t)

The integrating factor is e^(ln(t)) = t.

Step 2: Multiply the equation by the integrating factor.
t (t dQ/dt) + t(Q) = t^2 dQ/dt + tQ = t^5 ln(t)

Step 3: Integrate both sides of the equation.
∫(t^2 dQ/dt + tQ)dt = ∫(t^5 ln(t))dt

Using integration by parts on the right side (u = ln(t), dv = t^5 dt):

∫(t^5 ln(t))dt = (t^5 ln(t) / 5) - ∫(t^4 dt) = (t^5 ln(t) / 5) - (t^5 / 25) + C

Step 4: Solve for Q(t).
Since ∫(t^2 dQ/dt + tQ)dt = tQ, we have:

tQ = (t^5 ln(t) / 5) - (t^5 / 25) + C
Q(t) = -t^4/25 + t^4/5 ln(t) + C/t

So, the solution to the given differential equation is:

Q(t) = -t^4/25 + t^4/5 ln(t) + C/t

For more questions on differential equations -  https://brainly.com/question/1164377

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