The perimeter of a rectangle is 84 cm. if the length is 25 cm, how wide is the rectangle?

1 answer:

6 0

P = 2(L + W)
P = 84
L = 25
now sub

84 = 2(25 + W)
84 = 50 + 2W
84 - 50 = 2W
34 = 2W
34/2 = W
17 = W <=== the width is 17 cm

You might be interested in

Marry me if you can answer this-

horsena [70]

7 I think, I’m sorry if I’m wrong

5 0

3 years ago

Simplify<br><img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B%20%5Cfrac%7B3%7D%7B8%7D%20%7D%20" id="TexFormula1" title=" \sqrt{ \f

Pie

Answer:

B (√6)/4

Step-by-step explanation:

The smallest multiplier that will make the denominator of the fraction into a perfect square is 2, so you have ...

$\displaystyle\sqrt{\frac{3}{8}}=\sqrt{\frac{3\cdot 2}{8\cdot 2}}\\\\=\sqrt{\frac{6}{16}}=\frac{\sqrt{6}}{\sqrt{16}}\\\\=\frac{\sqrt{6}}{4}$

_____

Answer choice D is also a correct rationalization of the denominator, but is not simplified as far as it can be. √24 = 2√6, so a factor of 2 can be cancelled from numerator and denominator, giving answer choice B.

3 0

3 years ago

Anne is choosing a contractor. She has narrowed down her selection to two

MArishka [77]

Answer: the answer is contractor A: the probability that both conditions are met is 0.92.

Step-by-step explanation:

6 0

3 years ago

Please tell me the expression to this

kobusy [5.1K]

Answer:

5(5x+11)

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Find the distance between each pair of points. Round to the nearest tenth, if necessary.

Westkost [7]

$\bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
O&({{ 0}}\quad ,&{{ 0}})\quad 
% (c,d)
P&({{ -5}}\quad ,&{{ 6}})\\
K&({{ 5}}\quad ,&{{ 0}})\quad 
% (c,d)
L&({{ -2}}\quad ,&{{ 1}})

\end{array}\qquad 
% distance value
d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}
\\\\\\
OP=\sqrt{(-5-0)^2+(6-0)^2}\qquad \qquad KL=\sqrt{(-2-5)^2+(1-0)^2}$

3 0

3 years ago