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4 years ago

8

the costume designer for the school play uses 4 feet of fabric to make 8 costumes. At the same rate, how many costumes can the d

esigner make with 4 yards of fabric?

1 answer:

DiKsa [7]4 years ago

7 0

4 times 3=12 feet times two 24 costumes with 4 yards

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Find the inverse of each function for problems 1–6. State the domain and range of both the function and its inverse. Restrict th

JulsSmile [24]

Answer:

1.

<u>Function:</u>

$f(x)=-x^2$

Domain: (-∞,∞)

Range: (-∞,0]

<u>Inverse Function:</u>

$f^{-1}(x)=\sqrt{-x} ,and\\f^{-1}(x)=-\sqrt{-x}$

Domain: (-∞,0]

Range: (-∞,∞)

2.

<u>Function:</u>

$f(x)=5x-1$

Domain: (-∞,∞)

Range: (-∞,∞)

<u>Inverse Function:</u>

$f^{-1}(x)=\frac{1}{5}x+\frac{1}{5}$

Domain: (-∞,∞)

Range: (-∞,∞)

3.

<u>Function:</u>

$f(x)=-x+3$

Domain: (-∞,∞)

Range: (-∞,∞)

<u>Inverse Function:</u>

$f^{-1}(x)=-x+3$

Domain: (-∞,∞)

Range: (-∞,∞)

4.

<u>Function:</u>

$f(x)=x^{2}+7$

Domain: (-∞,∞)

Range: [7,∞)

<u>Inverse Function:</u>

$f^{-1}(x)=\sqrt{x-7}, and\\f^{-1}(x)=-\sqrt{x-7}$

Domain: [7,∞)

Range: (-∞,∞)

5.

<u>Function:</u>

$f(x)=14x-4$

Domain: (-∞,∞)

Range: (-∞,∞)

<u>Inverse Function:</u>

$f^{-1}(x)=\frac{1}{14}x+\frac{2}{7}$

Domain: (-∞,∞)

Range: (-∞,∞)

6.

<u>Function:</u>

$f(x)=-3x+8$

Domain: (-∞,∞)

Range: (-∞,∞)

<u>Inverse Function:</u>

$f^{-1}(x)=-\frac{1}{3}x+\frac{8}{3}$

Domain: (-∞,∞)

Range: (-∞,∞)

Step-by-step explanation:

To find inverse of a function f(x), there are 4 steps we need to follow:

1. Replace f(x) with y

2. Interchange the y and x

3. Solve for the "new" y

4. Replace the "new" y with the notation for inverse function, $f^{-1}(x)$

<u>Note:</u> The domain of the original function f(x) is the range of the inverse and the range of the original function is the domain of the inverse function.

<u><em>Let's calculate each of these.</em></u>

1.

$f(x)=-x^2$

Domain: There is no restriction on values of x we can put on it. Hence domain is (-∞,∞)

Range: No matter what we put into x, the y values will always be negative. And if we put 0, y value would be 0. So range is (-∞,0]

<u>Finding the inverse:</u>

$f(x)=-x^2\\y=-x^2\\x=-y^2\\y^2=-x\\y=+-\sqrt{-x} \\y=\sqrt{-x}, -\sqrt{-x}$

So

$f^{-1}(x)=\sqrt{-x} ,and\\f^{-1}(x)=-\sqrt{-x}$

Domain: this is the range of the original so domain is (-∞,0]

Range: this is the domain of the original so range is (-∞,∞)

2.

$f(x)=5x-1$

Domain: There is no restriction on values of x we can put on it. Hence domain is (-∞,∞)

Range: All sorts of y values will occur, so the range is (-∞,∞)

<u>Finding the inverse:</u>

$f(x)=5x-1\\y=5x-1\\x=5y-1\\5y=x+1\\y=\frac{1}{5}x+\frac{1}{5}$

So

$f^{-1}(x)=\frac{1}{5}x+\frac{1}{5}$

Domain: this is the range of the original so domain is (-∞,∞)

Range: this is the domain of the original so range is (-∞,∞)

3.

$f(x)=-x+3$

Domain: There is no restriction on values of x we can put on it. Hence domain is (-∞,∞)

Range: All sorts of y values will occur, so the range is (-∞,∞)

<u>Finding the inverse:</u>

$f(x)=-x+3\\y=-x+3\\x=-y+3\\y=-x+3$

So

$f^{-1}(x)=-x+3$

Domain: this is the range of the original so domain is (-∞,∞)

Range: this is the domain of the original so range is (-∞,∞)

4.

$f(x)=x^{2}+7$

Domain: There is no restriction on values of x we can put on it. Hence domain is (-∞,∞)

Range: no matter what we put into x, it will always be a positive number greater than 7. Only when we put in 0, y will be 7. So 7 is the lowest number and it can go to infinity. Hence the range is [7,∞)

<u>Finding the inverse:</u>

$f(x)=x^2+7\\y=x^2+7\\x=y^2+7\\y^2=x-7\\y=+-\sqrt{x-7}$

So

$f^{-1}(x)=\sqrt{x-7}, and\\f^{-1}(x)=-\sqrt{x-7}$

Domain: this is the range of the original so domain is [7,∞)

Range: this is the domain of the original so range is (-∞,∞)

5.

$f(x)=14x-4$

Domain: There is no restriction on values of x we can put on it. Hence domain is (-∞,∞)

Range: no matter what we put into x, we can get any y value from negative infinity to positive infinity. So range is (-∞,∞)

<u>Finding the inverse:</u>

$f(x)=14x-4\\y=14x-4\\x=14y-4\\14y=x+4\\y=\frac{1}{14}x+\frac{2}{7}$

So

$f^{-1}(x)=\frac{1}{14}x+\frac{2}{7}$

Domain: this is the range of the original so domain is (-∞,∞)

Range: this is the domain of the original so range is (-∞,∞)

6.

$f(x)=-3x+8$

Domain: There is no restriction on values of x we can put on it. Hence domain is (-∞,∞)

Range: no matter what we put into x, we can get any y value from negative infinity to positive infinity. So range is (-∞,∞)

<u>Finding the inverse:</u>

$f(x)=-3x+8\\y=-3x+8\\x=-3y+8\\3y=-x+8\\y=-\frac{1}{3}x+\frac{8}{3}$

So

$f^{-1}(x)=-\frac{1}{3}x+\frac{8}{3}$

Domain: this is the range of the original so domain is (-∞,∞)

Range: this is the domain of the original so range is (-∞,∞)

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4 years ago

Bill and Mofor each improved their yards by planting hostas and ivy. They bought their supplies from the same store. Bill spent

masha68 [24]

Answer: sorry I don’t kno

Step-by-step explanation:

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3 years ago

99 (567834 / 786) what is this answer

Zinaida [17]

Answer:

71521.0763359

Step-by-step explanation:

first solve the equation in the parentheses than multiply your answer by 99

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4 years ago

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laws of Sines with find the angle find each measurement indicated. Round your answer to the nearest tenth.

Kitty [74]

Answer:

B ≈ 34.0°

Step-by-step explanation:

We want to use the Law of Sines, which states that for a triangle with side lengths a, b, and c and angles A, B, and C:

$\frac{a}{sinA} =\frac{b}{sinB} =\frac{c}{sinC}$

Here, b = 14, c = 25, and C = 93, so plug these in:

$\frac{b}{sinB} =\frac{c}{sinC}$

$\frac{14}{sinB} =\frac{25}{sin(93)}$

Solve for B:

B ≈ 34.0°

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3 years ago