A rectangle has a length of 9 inches and a perimeter of 22 inches. What is the width of the rectangle?

Mathematics

1 answer:

Minchanka [31]3 years ago

7 0

$p = 2(l + w)$

$22 = 2(9 + w)$

$\frac{22}{2} = \frac{2(9 + w)}{2}$

$11 = 9 + w$

$11 - 9 = 9 - 9 + w$

$2 = w$

$p = 2(9 + 2)$

$p = 2(11)$

$p = 22$

You might be interested in

Y=x+5<br> 4x+y=20<br> Solving system by substitution

sineoko [7]

Y = x + 5....so we sub in x + 5 for y in the other equation

4x + y = 20
4x + x + 5 = 20
5x + 5 = 20
5x = 20 - 5
5x = 15
x = 15/5
x = 3

y = x + 5
y = 3 + 5
y = 8

solution is (3,8)

3 0

3 years ago

PLEASE HELP MEEE Instruction

natali 33 [55]

combine like terms on the left side. you always go from left to right.

5 0

3 years ago

Find the six trigonometric function values for angle ∅ where its adjacent side is -9 and its hypotenuse is 41. (Theta is located

arsen [322]

Check the picture below.

$\bf \textit{using the pythagorean theorem}
\\\\
c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b
\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}
\\\\\\
\sqrt{41^2-(-9)^2}=b\implies \sqrt{1681-81}=b\\\\\\ \sqrt{1600}=b\implies 40=b\\\\
-------------------------------$

$\bf sin(\theta )=\cfrac{\stackrel{opposite}{40}}{\stackrel{hypotenuse}{41}}\qquad~~ cos(\theta )=\cfrac{\stackrel{adjacent}{-9}}{\stackrel{hypotenuse}{41}}\qquad~~ tan(\theta )=\cfrac{\stackrel{opposite}{40}}{\stackrel{adjacent}{-9}}
\\\\\\
csc(\theta )=\cfrac{\stackrel{hypotenuse}{41}}{\stackrel{opposite}{40}}\qquad ~~sec(\theta )=\cfrac{\stackrel{hypotenuse}{41}}{\stackrel{adjacent}{-9}}\qquad ~~cot(\theta )=\cfrac{\stackrel{adjacent}{-9}}{\stackrel{opposite}{40}}$