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2 years ago

Solve for C. 4/ 7(B - C) = A C = ___

Mathematics

1 answer:

klio [65]2 years ago

5 0

Answer:

C = ¼(4B – 7A)

Step-by-step explanation:

⁴/₇(B – C) = A Multiply each side by 7

4(B - C) = 7A Remove parentheses

4B – 4C = 7A Subtract 4B from each side

-4C = 7A – 4B Multiply each side by -1

4C = -7A + 4B Divide each side by 4

C = ¼(4B – 7A)

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Wewaii [24]

Answer:

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Step-by-step explanation:

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2 years ago

The length of a rectangle is 2 centimeters more than five times the width. The perimeter is 114 inches. Find the width and the l

gogolik [260]

1. Let L=length and W=width
2. L=5w+2
3. The perimeter can be solved as follows:
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Answer:

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2 years ago

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Step-by-step explanation:

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2 years ago

The function f(x)=(x-5)^2 +2 is not one-to-one. Identify a restricted domain that makes the function one-to-one, and find the in

Anon25 [30]

We have been given a quadratic function $f(x)=(x-5)^{2} +2$ and we need to restrict the domain such that it becomes a one to one function.

We know that vertex of this quadratic function occurs at (5,2).

Further, we know that range of this function is $[2,\infty)$ .

If we restrict the domain of this function to either $(-\infty,5]$ or $[5,\infty)$ , it will become one to one function.

Let us know find its inverse.

$y=(x-5)^{2}+2$

Upon interchanging x and y, we get:

$x=(y-5)^{2}+2$

Let us now solve this function for y.

$(y-5)^{2}=x-2\\
y-5=\pm \sqrt{x-2}\\
y=5\pm \sqrt{x-2}\\$

Hence, the inverse function would be $f^{-1}(x)=5+\sqrt{x-2}$ if we restrict the domain of original function to $[5,\infty)$ and the inverse function would be $f^{-1}(x)=5-\sqrt{x-2}$ if we restrict the domain to $(-\infty,5]$ .

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2 years ago