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3 years ago

7

The rectangular sandbox at the local community park has a width of 24.5 meters and its length is 31.7 meters. What is the perime

ter, in meters, of the rectangular sandbox?

1 answer:

eduard3 years ago

5 0

The answer is 112.4 because 24.5+24.5+31.7+31.7=112.4.

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Please help quickly !

Ira Lisetskai [31]

Answer:

75 - 52.5 = x

Step-by-step explanation:

$\because 52.5 + x = 75\\\therefore x = 75 - 52.5\\\therefore 75 - 52.5 = x \\$

is the correct answer.

7 0

3 years ago

What ya the value of k? <br> A<br> B<br> C<br> D

sveta [45]

<h3>♫ - - - - - - - - - - - - - - - ~Hello There!~ - - - - - - - - - - - - - - - ♫</h3>

➷ Angles on a straight line equal 180 degrees:

5k - 3 + 9 + k = 180

Simplify:

6k + 6 = 180

Subtract 6 from both sides:

6k = 174

Divide both sides by 6:

k = 29

<h3><u>✽</u></h3>

➶ Hope This Helps You!

➶ Good Luck (:

➶ Have A Great Day ^-^

↬ ʜᴀɴɴᴀʜ ♡

7 0

4 years ago

1. Solve for x. Show each step of the solution<br> 4.5(8 - x) + 36 = 102 - 2.5(3x + 24)

Vladimir [108]

36-4.5x+36=102-7.5x-60
72-4.5x=42-7.5x
72+3.5x=42
3.5x=-30
x=8.57

7 0

3 years ago

Select each perfect square

3241004551 [841]

Answer:

hi

Step-by-step explanation:

1 and 2 no are perfect square other r not

8 0

3 years ago

Read 2 more answers

Two problems here I need solved! I need every step, so please have that with your answers!!

Free_Kalibri [48]

QUESTION 1

We want to solve,

$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x^{2}-6x+8}$

We factor the denominator of the fraction on the right hand side to get,

$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x^{2}-4x - 2x+8}.$

This implies
$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x(x-4) - 2(x - 4)}.$

$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{(x-4)(x - 2)}$

We multiply through by LCM of
$(x-4)(x - 2)$

$(x - 2) + x(x-4) = 2$

We expand to get,

$x - 2 + {x}^{2} - 4x= 2$

We group like terms and equate everything to zero,

${x}^{2} + x - 4x - 2 - 2 = 0$

We split the middle term,

${x}^{2} + - 3x - 4 = 0$

We factor to get,

${x}^{2} + x - 4x- 4 = 0$

$x(x + 1) - 4(x + 1) = 0$

$(x + 1)(x - 4) = 0$

$x + 1 = 0 \: or \: x - 4 = 0$

$x = - 1 \: or \: x = 4$

But
$x = 4$
is not in the domain of the given equation.

It is an extraneous solution.

$\therefore \: x = - 1$
is the only solution.

QUESTION 2

$\sqrt{x+11} -x=-1$

We add x to both sides,

$\sqrt{x+11} =x-1$

We square both sides,

$x + 11 = (x - 1)^{2}$

We expand to get,

$x + 11 = {x}^{2} - 2x + 1$

This implies,

${x}^{2} - 3x - 10 = 0$

We solve this quadratic equation by factorization,

${x}^{2} - 5x + 2x - 10 = 0$

$x(x - 5) + 2(x - 5) = 0$

$(x + 2)(x - 5) = 0$

$x + 2 = 0 \: or \: x - 5 = 0$

$x = - 2 \: or \: x = 5$

But
$x = - 2$
is an extraneous solution

$\therefore \: x = 5$

7 0

4 years ago