What is the slope of the line shown below? A. -1/4 B. 4 C. -4 D. 1/4

The length of a rectangle is eight more than twice its width. The perimeter is 96 feet. Find the dimensions of the rectangle

Debora [2.8K]

Answer:

The width of the rectangle is 14'8", and the length is 33'4"

Step-by-step explanation:

We're given two pieces of information:

The length is eight more than twice the width:

$l = 2w + 4$

The perimeter is 96 feet:

$p = 96$

We also need to apply one more piece of information that is not provided here, and that is the relationship between the perimeter of a rectangle, and it's length and width:

$p = 2l + 2w$

We can solve for w by plugging the other two values into the last:

$p = 2w + 2l\\96 = 2w + 2(2w + 4)\\96 = 2w + 4w + 8\\88 = 6w\\3w = 44\\w = \frac{44}{3}\\w = 14\frac{2}{3}$

Now we can find the length by plugging w into the first equation:

$l = 2w + 4\\l = 2(14\frac{2}{3}) + 4\\l = 29\frac{1}{3} + 4\\l = 33\frac{1}{3}$

One third of a foot is four inches, so the width is 14'8" and the length is 33'4"

To make sure our answer is correct, we should plug those numbers back into the area equation and see if we're right:

$p = 2l + 2w\\p = 2(33\frac{1}{3}) + 2(14\frac{2}{3})\\p = 66\frac{2}{3} + 29\frac{1}{3}\\p = 96$

So we know our answer's correct

8 0

2 years ago

Can you help me please and thank you

dezoksy [38]

Yeah there right it’s B

5 0

2 years ago

Find the tangent line approximation for 10+x−−−−−√ near x=0. Do not approximate any of the values in your formula when entering

Svetllana [295]

Answer:

$L(x)=\sqrt{10}+\frac{\sqrt{10}}{20}x$

Step-by-step explanation:

We are asked to find the tangent line approximation for $f(x)=\sqrt{10+x}$ near $x=0$ .

We will use linear approximation formula for a tangent line $L(x)$ of a function $f(x)$ at $x=a$ to solve our given problem.

$L(x)=f(a)+f'(a)(x-a)$

Let us find value of function at $x=0$ as:

$f(0)=\sqrt{10+x}=\sqrt{10+0}=\sqrt{10}$

Now, we will find derivative of given function as:

$f(x)=\sqrt{10+x}=(10+x)^{\frac{1}{2}}$

$f'(x)=\frac{d}{dx}((10+x)^{\frac{1}{2}})\cdot \frac{d}{dx}(10+x)$

$f'(x)=\frac{1}{2}(10+x)^{-\frac{1}{2}}\cdot 1$

$f'(x)=\frac{1}{2\sqrt{10+x}}$

Let us find derivative at $x=0$

$f'(0)=\frac{1}{2\sqrt{10+0}}=\frac{1}{2\sqrt{10}}$

Upon substituting our given values in linear approximation formula, we will get:

$L(x)=\sqrt{10}+\frac{1}{2\sqrt{10}}(x-0)$

$L(x)=\sqrt{10}+\frac{1}{2\sqrt{10}}x-0$

$L(x)=\sqrt{10}+\frac{\sqrt{10}}{20}x$

Therefore, our required tangent line for approximation would be $L(x)=\sqrt{10}+\frac{\sqrt{10}}{20}x$ .

8 0

2 years ago

What value of m makes the equation 8am/4a5=2a6 true?

skelet666 [1.2K]

Answer:

m = 11

Step-by-step explanation:

11 - 5 = 6

4 0

2 years ago

PLEASE HELP What is the x value??

Mrrafil [7]

Answer:

9

Step-by-step explanation:

21^2-19^2=80

sqaure root of 80 to the nearest 10th is

6 0

2 years ago