All categories

3 years ago

The length of a rectangle is 3 inches more than its width, and its perimeter is 22 inches. Find the width of the rectangle.

Mathematics

1 answer:

Natasha_Volkova [10]3 years ago

3 0

Answer:

W = 4 in

Step-by-step explanation:

Here W is the width and L the length. Then L = W + 3 in.

The formula for the perimeter of a rectangle is P = 2W + 2L.

Thus, 22 in = 2(W) + 2(W + 3), which reduces to

11 in = W + W + 3 in, or

8 in = 2W, and W = 4 in

You might be interested in

<img src="https://tex.z-dn.net/?f=if%20%5C%3A%20100%20%7B%20%5C%3A%20%20%7D%5E%7B2%7D%20%20%5C%3A%20%20%3D%20251p%20%5C%3A%20sol

Nikolay [14]

= 39.84
:
100^2=251
10000=251
ℎ
251 ℎ
10000/251=251/251
39.84=1
39.84=

8 0

3 years ago

Read 2 more answers

Which exponential equation is equation is equivalent to the logarithmic equation below? Log 200 = a

weqwewe [10]

Answer:

Which exponential equation is equation is equivalent to the logarithmic equation below? Log 200 = a

A) 200^10=a

B)a^10=200

C)200^a=10

D)a0^a=200

D) 10^a = 200 is the answer

Hope This Helps! Have A Nice Day!!

4 0

3 years ago

Read 2 more answers

What is 4×7×7×10- 5 in exponential form

liraira [26]

The answer is 977 seconds squared

7 0

3 years ago

The area of a rectangle is 52 m2 , and the length of the rectangle is 5 m less than double the width. Find the dimensions of the

kotegsom [21]

52m^2= 2[(x-5)+(x)]. 52m^2= 2(2x-5). 52m^2=4x-10

7 0

3 years ago

Find the radius of convergence, r, of the series. ∞ xn 2n − 1 n = 1 r = 1 find the interval, i, of convergence of the series. (e

Bingel [31]

Assuming the series is

$\displaystyle\sum_{n\ge1}\frac{x^n}{2n-1}$

The series will converge if

$\displaystyle\lim_{n\to\infty}\left|\frac{\frac{x^{n+1}}{2(n+1)-1}}{\frac{x^n}{2n-1}}\right|$

We have

$\displaystyle\lim_{n\to\infty}\left|\frac{\frac{x^{n+1}}{2(n+1)-1}}{\frac{x^n}{2n-1}}\right|=|x|\lim_{n\to\infty}\frac{\frac1{2n+1}}{\frac1{2n-1}}=|x|-\lim_{n\to\infty}\frac{2n-1}{2n+1}=|x|$

So the series will certainly converge if $-1$ , but we also need to check the endpoints of the interval.

If $x=1$ , then the series is a scaled harmonic series, which we know diverges.

On the other hand, if $x=-1$ , by the alternating series test we can show that the series converges, since

$\left|\dfrac{(-1)^n}{2n-1}\right|=\dfrac1{2n-1}\to0$

and is strictly decreasing.

So, the interval of convergence for the series is $-1\le x$ .

6 0

3 years ago