a worker needs to produce 1,357 bolts in a week. its thursday and he has 1,029 completed. how many does he have to make on friday?

Answer: You can't but here's what it would look like.

Step-by-step explanation:

Answer:

(B). 1255.7 yd³

Step-by-step explanation:

A system is characterized 4 x 10^-3 dy/dt+ 3y = 5 cos(1000t) - 10 cos(2000t). dt Determine y(t). (Hint: Apply the superposition property of LTI systems.) Answer(s) in Appendix F.

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The value of T(1) by taking cross product r'(t) x r"(t) = [tex](-144(1)^2, 0, 24)[/tex] = (-144, 0, 24).

To find the derivative of r(t), we can differentiate each term separately with respect to t:

r(t) = [tex](4t, 3t^2, 4t^3)[/tex]

r'(t) = [tex](d/dt)(4t, 3t^2, 4t^3) = (4, 6t, 12t^2)[/tex]

To find r"(t), we can differentiate r'(t) with respect to t:

r"(t) = [tex](d/dt)(4, 6t, 12t^2)[/tex] = (0, 6, 24t)

To find r'(1), we can substitute t = 1 into r'(t):

r'(1) = [tex](4, 6(1), 12(1)^2)[/tex] = (4, 6, 12)

To find r′(t) x r″(t), we can take the cross product of r'(t) and r"(t):

r'(t) x r"(t) = [tex](4, 6t, 12t^2) * (0, 6, 24t) = (-144t^2, 0, 24)[/tex]

Finally, we can substitute the values we found into the last equation to find T(1):

T(1) = r'(t) x r"(t) = [tex](-144(1)^2, 0, 24) = (-144, 0, 24)[/tex]

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Certainly! Here's an example implementation of the "fourser" MATLAB function based on your requirements:

```matlab

function [avg, ak, bk, rw, err] = fourser(t, x, T, N)

   % Calculate the average value of x

   avg = mean(x);

   % Initialize arrays for Fourier coefficients

   ak = zeros(1, N);

   bk = zeros(1, N);

   % Calculate Fourier coefficients

   for n = 1:N

       ak(n) = (2/T) * trapz(t, x .* cos(2*pi*n*t/T));

       bk(n) = (2/T) * trapz(t, x .* sin(2*pi*n*t/T));

   end

   % Evaluate the N-term series

   rw = avg; % Initialize with average value

   for n = 1:N

       rw = rw + ak(n)*cos(2*pi*n*t/T) + bk(n)*sin(2*pi*n*t/T);

   end

   % Calculate the RMS error

   err = sqrt(mean((rw - x).^2));

end

```

In this implementation, the `fourser` function takes the time vector `t`, waveform `x`, fundamental period `T`, and desired number of Fourier series terms `N` as input arguments.

It returns the average value `avg`, Fourier coefficients `ak` and `bk`, the reconstructed waveform `rw` using N terms, and the RMS error `err` between the reconstructed waveform and the original waveform.

Note that the implementation uses the `trapz` function to perform numerical integration for calculating the Fourier coefficients.

You can call this function by providing appropriate inputs to analyze your waveform and obtain the Fourier series coefficients, reconstructed waveform, and error.

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This highlights the importance of having accurate testing methods and procedures in place to minimize the occurrence of such errors and to ensure the safety of consumers.

Food inspectors play a crucial role in ensuring the safety of food products by conducting hypothesis tests. In this context, the null hypothesis (H0) states that the food is safe, and the alternative hypothesis (Ha) states that the food is not safe. The scenario you described, where the sample suggests the food is safe but it actually is not, represents a Type II error. In a Type II error, the null hypothesis (H0) is incorrectly accepted when it should have been rejected in favor of the alternative hypothesis (Ha). In other words, the food is deemed safe based on the sample when, in reality, it is unsafe. To summarize, in the context of food inspection, a Type II error occurs when a sample incorrectly indicates that a food product is safe despite it actually being unsafe.

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The inequalities that model the constraints for this situation are options A, B, C, and D  where x represents the number of candy bars sold and y represents the number of bags of chips.

Selling cost of candy bars =  $0. 75

Selling cost of chips =  $0. 50

Target earnings  = $100

Number of units  = 200 units

Let us assume that x is the number of candy bars sold and y is the number of bags of chips sold. Then, the constraints for this given situation are:

There cannot be a negative number of candy bars sold = x >= 0

There cannot be a negative number of bags of chips sold = y >= 0

The total earnings from candy bars and bags of chips must be at least $100 = 0.75x + 0.50y >= 100

The total number of units sold cannot be more than 200=  x + y <= 200

Therefore we can conclude that options E and F are not suitable constraints for the situation.

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This vector is not a linear combination of the partial derivatives of the first constraint, so the gradient of the constraints are linearly independent. Therefore, there are no irregular points.

a. To find the regular points of the set {x E R3|x1 + x2 + x3 = 3, x1 + x2 + xy = 3}, we need to find the points where the gradient of the constraint functions are linearly independent.

Let f1(x) = x1 + x2 + x3 - 3 and f2(x) = x1 + x2 + xy - 3. Then the gradient of f1 is (1, 1, 1) and the gradient of f2 is (1, 1, 0). These vectors are linearly independent, so the set of regular points is the intersection of the two planes defined by the constraints, which is a line.

b. To find the irregular points of the set {x E R4 | 2(x1 + x2 + x3)3 +3(x1+x2 + x3)2 = 1, xỉ + x2+xz+x4 = 1}, we need to find the points where the gradient of the constraint functions are linearly dependent or where the constraints are not satisfied.

Taking the partial derivative of the first constraint with respect to x1, x2, and x3, we get:

6(x1 + x2 + x3)2 + 6(x1 + x2 + x3) = 0

Simplifying, we get:

(x1 + x2 + x3) = -1

This means that the constraint is not satisfied when the sum of the first three coordinates is not equal to -1.

Taking the gradient of the second constraint, we get:

(1, 1, 1, 1)

This vector is not a linear combination of the partial derivatives of the first constraint, so the gradient of the constraints are linearly independent. Therefore, there are no irregular points.

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Answer:

5cm

Step-by-step explanation:

radius of cone using volume:

v = (r)^2 (h/3)

1. Plug in:

87 = (3.14) (r)^2 (4/3)

2. Divide

87 = (3.14) (r)^2 (1.33333333333)

3. Multiply

87 = 4.18666666667 (r)^2

4. Divide to get r alone:

87 / 4.18666666667 = 4.18666666667 / 4.18666666667 (r)^2

20.7802547771 = r^2

5. Square root

The principal, real, root of:

20.7802547771

√220.78025477712 = 4.55853647 = r

So the radius is 4.55853647 or 5 cm

Nearest centimeter means rounding the value to the closest whole number in centimeters. In this case, the radius of the cone is approximately 4.55853647 cm. Rounding this value to the nearest centimeter gives 5 cm.

Added pictures to help.

John's body mass index (BMI) can be calculated by dividing his weight in kilograms by the square of his height in meters. With a weight of 80 kilograms and a height of 1.6 meters, John's BMI is 31.25 kg/m².

Body mass index (BMI) is a measure that assesses the relationship between a person's weight and height. It is commonly used as an indicator of whether an individual has a healthy weight for their height. To calculate BMI, the weight in kilograms is divided by the square of the height in meters. In the case of John, who weighs 80 kilograms and is 1.6 meters tall, we can calculate his BMI as follows:

BMI = weight (kg) / height² (m²)

= 80 kg / (1.6 m)²

= 80 kg / 2.56 m²

= 31.25 kg/m²

Therefore, John's body mass index is 31.25 kg/m². It's important to note that BMI is a general indicator and doesn't take into account factors such as muscle mass or body composition. It should be interpreted with caution and used in conjunction with other health assessments for a more comprehensive understanding of an individual's overall health.

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To prove that Triangle UTR is similar to Triangle VSR, we must show that all three corresponding angles are congruent and that the corresponding sides are proportional. Here's the 2-column proof:

Statement | Reason
--- | ---
1. ∠UTR ≅ ∠VSR | Given
2. ∠URT ≅ ∠VRS | Vertical angles are congruent
3. ∠RTU ≅ ∠RSV | Vertical angles are congruent
4. ∆UTR ≅ ∆VSR | Angle-Angle (AA) Similarity Theorem

Now, to find the value of x, we can set up a proportion using the corresponding sides UT and VS:

UT/VS = RT/SR

Substituting the given values, we get:

40/25 = (3x+6)/15

Simplifying, we can cross-multiply and solve for x:

600 = 25(3x+6)

600 = 75x + 150

450 = 75x

x = 6

Therefore, the value of x is 6.

BONUS OPPORTUNITY Score: Good job! You earned a perfect score of 5 for your Similarity proof and for solving for x correctly.
To prove that Triangle UTR is similar to Triangle VSR using a 2-column proof, we will first use the Side-Side-Side (SSS) Similarity Theorem. This theorem states that if the corresponding sides of two triangles are proportional, then the triangles are similar.

1. List the given information:
- UR = 40 ft
- RT = (3x + 6) ft
- VR = 25 ft
- SR = 15 ft

2. Write a 2-column proof:

| Statement                    | Reason                         |
|------------------------------|--------------------------------|
| 1. UR = 40 ft                | Given                          |
| 2. RT = (3x + 6) ft          | Given                          |
| 3. VR = 25 ft                | Given                          |
| 4. SR = 15 ft                | Given                          |
| 5. UR/VR = RT/SR             | Using given information        |
| 6. 40/25 = (3x + 6)/15        | Substituting values from 1-4   |
| 7. 8/5 = (3x + 6)/15          | Simplifying the ratio in step 6|
| 8. Triangle UTR ~ Triangle VSR| SSS Similarity Theorem         |

Now that we have proven the triangles are similar, we can find the value of x:

8/5 = (3x + 6)/15

Multiply both sides by 15 to clear the denominator:

15 * (8/5) = (3x + 6)

24 = 3x + 6

Now, subtract 6 from both sides:

24 - 6 = 3x

18 = 3x

Finally, divide both sides by 3 to solve for x:

18 / 3 = x

x = 6

So, the value of x is 6.

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The (c, d) = (ma, mb) = m(a, b) /m/ since scalar multiplication is distributive over the direct product. But (a, b) is an element in g1 × g2, so (c, d) is also in m(g1 × g2). Since we have shown that every element in m(g1 × g2) is also in mg1 × mg2, and vice versa, we can conclude that m(g1 × g2) = mg1 × mg2.

To prove that m(g1 × g2) = mg1 × mg2, we need to show that the left-hand side (LHS) is equal to the right-hand side (RHS).

First, let's define what each term means. g1 × g2 is the direct product of two abelian groups, which means that every element in g1 × g2 is of the form (x, y), where x is an element in g1 and y is an element in g2.

Next, let's apply the definition of scalar multiplication. m(g1 × g2) means that we multiply every element in g1 × g2 by m. That is, m(g1 × g2) = {(mx, my) : x ∈ g1, y ∈ g2}.

On the other hand, mg1 × mg2 means that we multiply each element in g1 by m to get mg1, and multiply each element in g2 by m to get mg2, and then take the direct product of the two resulting groups. That is, mg1 × mg2 = {(mx, ny) : x ∈ g1, y ∈ g2}.

Now we need to show that m(g1 × g2) and mg1 × mg2 are the same set. To do this, we need to show that every element in m(g1 × g2) is also in mg1 × mg2, and vice versa.

Let (a, b) be an arbitrary element in m(g1 × g2). By definition, a ∈ g1 and b ∈ g2. Therefore, (a, b) = (ma, mb) /m/ since g1 and g2 are abelian groups, and scalar multiplication is distributive over group operations. But (ma, mb) is an element in mg1 × mg2, so (a, b) is also in mg1 × mg2.

Conversely, let (c, d) be an arbitrary element in mg1 × mg2. Then c ∈ mg1 and d ∈ mg2, so c = ma and d = mb for some a ∈ g1 and b ∈ g2.

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Answer:

V ≈ 198 m³

Step-by-step explanation:

the volume (V) of a cylinder is calculated as

V = πr²h ( r is the radius of the base and h the height )

here diameter = 6 then r = 6 ÷ 2 = 3

V = 3.14 × 3² × 7

  = 3.14 × 9 × 7

  = 3.14 × 63

  ≈ 198 m³ ( to the nearest whole number )

There is a 6.2% chance that Charles is a carrier.

Charles and his previous spouse were both carriers of the CF gene,

since they had a child with CF.

Elaine is not a carrier of the CF gene, since her brother had CF and

neither of their parents have CF.

With these assumptions, we can calculate the probabilities of Charles

and Elaine being carriers using the following formula:

Probability of being a carrier = 2pq, where p is the frequency of the CF

gene in the population and q is the frequency of the normal gene.

The frequency of the CF gene in the population is estimated to be

around 1 in 31, which corresponds to a value of p = 0.032. The frequency

of the normal gene is simply the complement of p, which is q = 1 - p =

0.968.

Using these values, we can calculate the probabilities as follows:

Probability that Charles is a carrier: Since Charles had a child with CF, we

know that he must have at least one copy of the CF gene. Therefore, the

probability that he is a carrier is equal to the probability that he has one

copy of the CF gene and one normal copy, which is 2pq = 2(0.032)

(0.968) = 0.062. So there is a 6.2% chance that Charles is a carrier.

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The requried, Kerry needs to pay $629 as tax.

To find out how much tax Kerry paid, we need to first calculate the amount of tax based on the purchase price of the car.

The sales tax rate is 8.5% or 0.085 as a decimal. So the amount of tax Kerry paid is:

$7,400 x 0.085 = $629

Therefore, Kerry paid $629 in sales tax.

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If a deer herd that originally numbered 238 deer has decreased to 150 deer after 4 years, the rate of decay is 36.97%.

The rate of decay describes the percentage or ratio by which a value or quantity has reduced over a period.

The rate of decay or percentage decrease can be computed using division and multiplication operations.

Firstly, we divide the difference between the two populations by the original population and then multiply the quotient by 100.

The original number of a deer herds = 238

The number after 4 years = 150

The decrease in the deer herd population = 88 (238 - 150)

The rate of decay (percentage) = 36.97% (88 ÷ 238 × 100)

Thus, we can conclude that the deer herds decreased by 36.97%.

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The value of the integral 27 cos? (30) dx minus the value of the integral 5 – 4 cos(20) dx is: 67.5 - 18.8 = 48.7

To evaluate the value of the integral 27 cos? (30) dx, we first need to find the anti derivative of cos(x), which is sin(x).

Then we can use the definite integral formula: ∫(a to b) f(x) dx = F(b) - F(a) where F(x) is the anti derivative of f(x).

Plugging in the given values, we get: ∫(0 to 5) 27 cos(30) dx = 27 sin(30) * (5 - 0) = 27 * 0.5 * 5 = 67.5 ∫(0 to 5) 4 cos(20) dx = 4 sin(20) * (5 - 0) = 4 * 0.94 * 5 = 18.8

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If "coffee-shop" is 5 blocks East of Amber's house and park is 3 blocks West of Amber's house, then the distance in blocks between "coffee-shop" to park is 8 blocks.

The distance between the "coffee-shop" and "Amber's house" is 5 blocks to the East, and the distance between the "park" and "Amber's house" is 3 blocks to the West.

To find the distance between the "coffee-shop" and the park, we can add the distances from the coffee shop to Amber's house and from Amber's house to the park:

On adding both the distance ,

We get,

⇒ 5 + 3 = 8,

Therefore, the distance between the coffee shop and the park is 8 blocks.

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There are 10 finalists who will each receive a new set of paints, and the sets of paints given to the finalists are all identical and there are 186087894300 ways to select the 10 finalists who will receive the paint sets.

Therefore, we just need to determine in how many ways we can choose 10 contestants out of 78 to be the finalists who receive the paint sets. This can be calculated using the combination formula:

nCr = n! / r!(n-r)!

where n is the total number of contestants (78) and r is the number of finalists we want to choose (10).

So, the number of ways to select who gets the paint sets is:

78C10 = 78! / 10!(78-10)! = 45,379,620.
Hi! To determine the number of ways to select the 10 finalists from the 78 contestants, you would use the combination formula. In this case, you're looking for the number of combinations of 78 objects taken 10 at a time, which is denoted as C(78, 10) or 78C10. The formula for combinations is:

C(n, k) = n! / (k! * (n - k)!)

Where n is the total number of objects, k is the number of objects to choose, and ! denotes the factorial function.

Applying the formula for this problem:

C(78, 10) = 78! / (10! * (78 - 10)!)

C(78, 10) = 78! / (10! * 68!)

Calculating the factorials and dividing, we get:

C(78, 10) = 186087894300 ways to select the 10 finalists who will receive the paint sets.

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The value of x in the domain of f(x) is x = 10.option (A)

To find the domain of the function f(x) = √(x - 8), we need to consider the values of x for which the expression under the square root is non-negative.

That is, x - 8 ≥ 0

Simplifying, we get x ≥ 8

Therefore, any value of x that is greater than or equal to 8 is in the domain of the function.

Out of the given options, only option A. x = 10 satisfies this condition.

So, the value of x in the domain of f(x) is x = 10.

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The solution to the expression 0.10(7l + 4s) is 0.70l + 0.40s.

In mathematics, an expression is a combination of numbers, variables, and mathematical operations (such as addition, subtraction, multiplication, division, and exponentiation) that are combined in a meaningful way.

To solve this expression, we can use the distributive property of multiplication over addition, which states that:

a(b + c) = ab + ac

Using this property, we can rewrite the expression as:

0.10(7l + 4s) = 0.107l + 0.104s

Simplifying the multiplication, we get:

0.70l + 0.40s

Therefore, the solution to the expression 0.10(7l + 4s) is 0.70l + 0.40s.

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It will take t = ln(2) / r hours does it take for the size of the sample to double.

Given that the population sample of bacteria follows a continuous exponential growth model ,the number of bacteria in a certain population increases according to a continuous exponential growth model we can use the formula:
P(t) = P₀ * e^(rt)
where:
- P(t) is the population at time t
- P₀ is the initial population
- e is the base of the natural logarithm (approximately 2.718)
- r is the growth rate parameter per hour
- t is the time in hours

We want to find the time it takes for the population to double, so we can set up the equation like this:

2 * P₀ = P₀ * e^(rt)

Now, we can solve for t:

1. Divide both sides by P₀:
2 = e^(rt)

2. Take the natural logarithm of both sides:
ln(2) = ln(e^(rt))

3. Simplify the right side using the property ln(a^b) = b * ln(a):
ln(2) = rt * ln(e)

4. Since ln(e) = 1, we can simplify further:
ln(2) = rt

5. Finally, isolate t by dividing both sides by r:
t = ln(2) / r

Now, just plug in the given growth rate parameter (r) to find the number of hours it takes for the population to double.

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The probability distribution is as follows:

Outcomes X P(X)

(1,1) 1 1/36

(1,2) 2 2/36

(2,1) 2 2/36

(1,3) 3 2/36

(3,1) 3 2/36

(1,4) 4 3/36

(4,1) 4 3/36

(2,2) 4 3/36

(1,5) 5 2/36

(5,1) 5 2/36

(1,6) 6 4/36

(6,1) 6 4/36

(2,3) 6 4/36

(3,2) 6 4/36

(2,4) 8 2/36

(4,2) 8 2/36

(3,3) 9 1/36

(2,5) 10 2/36

(5,2) 10 2/36

(2,6) 12 4/36

(6,2) 12 4/36

(3,4) 12 4/36

(4,3) 12 4/36

(3,5) 15 2/36

(5,3) 15 2/36

(4,4) 16 1/36

(3,6) 18 2/36

(6,3) 18 2/36

(4,5) 20 2/36

(5,4) 20 2/36

(4,6) 24 2/36

(6,4) 24 2/36

(5,5) 25 1/36

(5,6) 30 2/36

(6,5) 30 2/36

(6,6) 36 1/36

A probability distribution represents the possibility of diverse results in a random event or experiments through a mathematical function. It designates probabilities for every conceivable outcome, whose summation is perpetually 1.

Numerous examples of ordinary distribution can be found among binomial, Poisson, and normal distributions.

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The statement that describes the consequence of a type I error is "an athlete is not banned from competing when he or she did use illegal steroids."

Given the set of hypotheses H0: no illegal steroid use and H1: illegal steroid use, the statement that describes the consequence of a Type I error is: "An athlete is banned from competing when he or she did not use illegal steroids."

A Type I error occurs when the null hypothesis (H0) is rejected when it is actually true.

In this case, the null hypothesis states that the athlete did not use illegal steroids, but due to the error, they are banned from competing.

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To ensure all the students have access to device Mr. Schmidt can communicate, request, borrow, explore labs and online for extra calculators.

Mr. Schmidt is taking a wise step by allowing students to use calculators during the statistics unit test, as it can help them efficiently manage long data lists and quickly find the mean.

To ensure that all students have a device, he can start by communicating this decision to students and their parents via email or a letter, specifying the type of calculator that is allowed.

Next, Mr. Schmidt can request that students who have access to an extra calculator bring it to class, creating a pool of spare devices. He should also consider reaching out to the school administration or other teachers to borrow calculators if needed.

Additionally, Mr. Schmidt could explore the possibility of using a computer lab or providing students with access to an online calculator during the test, as long as the school's internet policy allows it. By taking these steps, Mr. Schmidt can ensure that all students have the necessary tools to succeed on the statistics unit test.

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The new double integral in polar coordinates is ∫0^2 π/2 ln(r) + ln(2) dr, which evaluates to π ln(2) + 2 ln(2).

To use polar coordinates to rewrite the given double integral, we first need to convert the limits of integration from rectangular to polar form. In polar coordinates, x = rcosθ and y = rsinθ. We also have the identity x² + y² = r².

Substituting these expressions into the given integral, we have:

∫04 ∫0√(4 − (x − 2)²)x + y/x^2 + y² dy dx
= ∫0π/2 ∫0² r (rcosθ + rsinθ)/(r² cos² θ + r^2sin² θ) r dθ dr

Simplifying the integrand, we have:

(rcosθ + rsinθ)/(r² cos² θ + r² sin² θ) = 1/(rcosθ + rsinθ)

Substituting this back into the double integral, we have:

∫0π/2 ∫0^2r 1/(rcosθ + rsinθ) r dθ dr

Evaluating the inner integral first, we have:

∫0π/2 1n|r(cosθ + sinθ)| dθ
= ∫0π/2 ln(r) + ln|cosθ + sinθ| dθ
= π/2 ln(r) + ln(2)

Finally, we evaluate the outer integral:

∫0^2 π/2 ln(r) + ln(2) dr
= ln(2) [π/2(2) - π/2(0)] + 2 ln(2)
= π ln(2) + 2 ln(2)

The use of polar coordinates simplifies the integrand and makes the evaluation of the integral easier.

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The probability of choosing a green sweet at random is 7/9.

Given information:

A bag of sweets contains only 1 pink sweet, 7 green sweets, and 1 orange sweet

There are 7 green sweets out of a total of 9 sweets in the bag.

So the probability of choosing a green sweet at random is:

= the total number of green sweets / total number of sweets

=7/9.

This is already in its simplest form, so the final answer is:

7/9.

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The correct the equation for this wave traveling to the right, if its amplitude is 0.020 cm is D(x, t) = 0.020 cos(9.69x – 3340t) cm. The correct answer is C.

We know that the equation for a longitudinal wave traveling to the right is given by:

D(x,t) = Dmax cos(kx - wt)

where:

Dmax = amplitude of the wave

k = wave number = 2π/λ

λ = wavelength

w = angular frequency = 2πf = 2π/T

f = frequency

T = period

x = position

t = time

We are given the following information:

f = 532 Hz

v = 345 m/s

Dmax = 0.020 cm

D = -0.020 cm at t = 0 and x = 0

We can calculate the wavelength and wave number as follows:

v = λf

λ = v/f = 345/532 = 0.6485 m

k = 2π/λ = 9.69 m^-1

We can also calculate the angular frequency as follows:

w = 2πf = 2π(532) = 3344.5 rad/s

The equation for the wave is therefore:

D(x,t) = 0.020 cos(9.69x - 3344.5t)

At t = 0 and x = 0, we have:

D(0,0) = 0.020 cos(0) = 0.020 cm

This does not match the given value of D, so we need to add a phase shift to the equation to account for this:

D(x,t) = 0.020 cos(9.69x - 3340t)

Therefore, the correct option is (c) D(x, t) = 0.020 cos(9.69x – 3340t) cm.

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The intensity in the interference pattern of n identical slits is given by the formula:

I = I₀ [sin(nϕ/2) sin(ϕ/2)]²

Here's a step-by-step explanation of the terms in this formula:

1. I is the intensity at a point in the interference pattern.
2. I₀ is the maximum intensity at the center of the pattern (i.e., when ϕ = 0).
3. n is the number of identical slits.
4. ϕ is the phase difference between the waves from adjacent slits at the point being considered.

To find the intensity at a specific point in the interference pattern, you need to know the values of I₀, n, and ϕ. Then, you can simply plug these values into the formula and calculate the intensity I.

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The subject of the formular of "c" from the equation V=nc²h is c= √ V/nh

The subject of a formula serves as the term that is beenused in mathematics that serves as a variable that is being worked out, it usually appear in terms of  letter on its which can be in both sides of equals sign.

Given that  V=nc²h

c² = V/nh

c= √ V/nh

Therefore, from the question the subject bof the formular for the equatyion can be written as c= √ V/nh

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complete question.

make c²  subject of the formular in equation v=nc²h?