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zzz [600]
3 years ago
14

A local drive in theatre sells entrance tickets for cars on weekends night. at capacity , the drive in can hold 72 cars . the fu

nction c(n)=25n represent the drive in's revenue on a weekend night, where n represents the number of cars. in this situation, what is the domain of c(n)?
Mathematics
2 answers:
amm18123 years ago
6 0

Answer:

0 ≤ n ≤ 72 where n is an integer

Step-by-step explanation:

Given : C(n) = 25n

To Find: what is the domain of c(n)?

Solution:

Domain of the function is defined as the values of n which we can use to substitute in the equation

Now, in  equation  C(n) = 25n , n represents the number of cars.

This means that n has to be positive integer because we can neither have negative number of cars nor fraction of a car.

This means , n would start from 0 .

The second thing is that the maximum capacity is 72. This means that n cannot exceed 72 cars in the drive-in

So, we can say that n can be any integer starting from 0 till 72 (inclusive)

Hence our domain is: 0 ≤ n ≤ 72 where n is an integer

Nitella [24]3 years ago
3 0
Its c

dont trust all the "its d"answers.  i got it wrong
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In a single growing season at the smith family orchard, the average yield per apple tree is 150 apples when the number of trees
Mariana [72]

Answer:

100 trees should be planted per acre to maximize the total yield (15000 apples per acre)

Step-by-step explanation:

<u>Functions </u>

Expressing some quantities in terms of other(s) comes in handy to understand the behavior of the situations often modeled by those functions.

Our problem tells us about the smith family orchard, where the average yield per apple tree is 150 apples when the number of trees per acre is 100 (or less). But the average yield per tree decreases when more than 100 trees are planted. The rate of decrease is 1 for each additional tree is planted

Let's call x the number of trees planted per acre in a single growing season. We have two conditions when modeling:  when x is less than 100, the average yield per apple tree (Ay) is (fixed) 150.

Ay(x)=150\ ,\ x\leqslant 100

When x is more than 100:

Ay(x)=150-x\ ,\ 100< x\leqslant 150

The upper limit is 150 because it would mean the Average yield is zero and no apples are to be harvested

Part 1

We are asked what happens if the number of trees per acre was doubled, but we don't know what the original number was. Let's make it x=60

If x=60, then we must use the first function, and

Ay(60)=150

The yield per acre will be 150*60=9000 apples

If we doubled x to 120, we would have to use the second function

Ay(120)=150-x=30

The yield per acre will be 30*120=3600 apples

Note that doubling the number of trees would make the production decrease

If we started with x=30, then  

Ay(30)=150, and we would have 150*30=4500 apples per acre

We already know that for x=60 (the double of 30), our production will be 9000 apples per acre

This shows us that in some cases, increasing the number of trees is beneficial and in other cases, it is not

Part 2

The total yield is computed as

Y(x)=x.Ay(x)

This function will be different depending on the value of x

Y(x)=150x\ ,\ x\leqslant 100

Y(x)=150x-x^2\ ,\ 100< x\leqslant 150

To find the maximum total yield we take the first derivative

Y'(x)=150\ ,\ x\leqslant 100

Y'(x)=150-2x\ ,\ 100< x\leqslant 150

The first expression has no variable, so the total yield must be evaluated in the endpoints

Y(0)=0

Y(100)=15000

Now, the second expression contains variable, it will be set to 0 to find the critical point

150-2x=0

x=75

x is outside of the boundaries, we cannot find the maximum in this interval

Answer: 100 trees should be planted per acre to maximize the total yield (15000 apples per acre)

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3 years ago
Match each division expression with the correct quotient ​
UkoKoshka [18]

Answer:

2) 2x + 6

3) 8x - 2

4) 5x - 8

Step-by-step explanation:

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3 years ago
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Find all values of the angle θ (in radians, with 0 ≤ θ &lt; 2π) for which the matrix a = cos θ −sin θ sin θ cos θ has real eigen
harina [27]

The matrix

A=\begin{bmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{bmatrix}

has eigenvalues \lambda such that

\det(A-\lambda I)=\begin{vmatrix}\cos\theta-\lambda&-\sin\theta\\\sin\theta&\cos\theta-\lambda\end{vmatrix}=0

(\cos\theta-\lambda)^2+\sin^2\theta=0

(\cos\theta-\lambda)^2=-\sin^2\theta

\cos\theta-\lambda=\pm\sqrt{-\sin^2\theta}

\lambda=\cos\theta\pm\sqrt{-\sin^2\theta}

\sin^2\theta\ge0 for all values of \theta, so we need to have \sin\theta=0 in order for \lambda to be real-valued. This happens for

\sin\theta=0\implies\theta=n\pi

where n is any integer, and over the given interval we have \theta=0 and \theta=\pi.

4 0
3 years ago
Find the area of the triangle with A=60º b=11ft and c=8ft
Serhud [2]

Answer:

38.105

Step-by-step explanation:

The area of this triangle  = 0.5 * b * c * sin(A)

= 0.5 * 11 * 8 *sin(60)           sin(60) = \sqrt{3} /2 = .866025...

= 0.5 * 11 * 8 * .866025

= 38.105

8 0
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A. Rewrite the function y = x2 – 14x + 58 in vertex form by completing the square. Show your work. B. Rewrite the function y = –
marysya [2.9K]

Answer:

A. y = (x -7)² +9

B. y = -(x +3)² -11

C. minimum, (7, 9)

D. maximum, (-3, -11)

Step-by-step explanation:

The general steps to rewriting the function in vertex form are ...

1. factor out the leading coefficient (if not 1) from the first two terms

2. add inside parentheses the square of half the x coefficient; subtract the same amount outside parentheses. The factor in front of the parentheses must be taken into account when adding the same amount outside.

3. rewrite the content of parentheses as a square, collect terms outside parentheses.

___

A. y = x² -14x +58

y = (x² -14x) +58 . . . . . . . . . . step 1

y = (x² -14x +49) +58 -49 . . . step 2

y = (x -7)² +9 . . . . . . . . . . . . . step 3

__

B. y = -x² -6x -20

y = -(x² +6x) -20

y = -(x² +6x +9) -20 -(-9)

y = -(x +3)² -11

__

C. In vertex form, the function for vertical scale factor "a" and vertex (h, k) looks like ...

y = a(x -h)² +k

When a < 0, the vertex is a maximum (the graph opens downward); when a > 0, the vertex is a minimum (the graph opens upward).

The function of part A has a minimum at the vertex (7, 9). 9 is the minimum value.

___

D. The function of part B has a maximum at the vertex (-3, -11). -11 is the maximum value.

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3 years ago
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