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2 years ago

Two similar parallelograms have areas of 18in^2 and 32in^2. find the ratio of their perimeters

1 answer:

3 0

$\displaystyle\\
k = \frac{P_1}{P_2} = \sqrt{ \frac{A_1}{A_2} }=\sqrt{\frac{18}{32}}=\sqrt{\frac{9\times2}{16\times2}}=\sqrt{\frac{9}{16}}=\frac{\sqrt{9}}{\sqrt{16}}= \boxed{\bf \frac{3}{4}}$

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The symbol ∪ represents the union of events. The union of two events A and B results in an event that contains all the sample po

Degger [83]

Answer:

AUB = {E1, E2, E4, E6, E7}

Step-by-step explanation:

Since we are to look for the union of the two sets, therefore we combine all the sample points that are contained in A and B without repeating any sample even if such sample occur in both sets. For example, E4 occurs in event A and B and as such will only be repeated once in the union of the two events.

AUB = {E1, E2, E4, E6, E7}

Note that the way you arrange does not matter as well.

I hope you find this solution useful?

8 0

2 years ago

How would you guys work this problem out

Effectus [21]

A = 5,460B = 5,733C = 7,280D = 6,720E = 8,050

8 0

2 years ago

Read 2 more answers

You purchase 5 pounds of cheese for $28. How much would 1 quart of cheese cost?

Vikentia [17]

Answer:

You would get 1.4

Step-by-step explanation:

First you divide 5 pounds by $28 your answer will be : 5.6

Than you divide 5.6 by 4 because 4 quarters make 1 whole

So, you divide 5.6 by 4

Than you get the answer 1.4.

7 0

2 years ago

Which of the following could not be a value of x? a -10 b-9 c-10.5 d-9.5

maw [93]

We know that sum of two sides in any triangle must be greater than the measure of third (longest) side otherwise triangle will not be formed.

According to picture longest side is 18

So this triangle must follow inequality

9+x>18

or 9-9+x>18-9

or x>9

that means value of x must be more than 9.

choices a, c, d are greater than 9 except choice b

Hence choice B-9 is the final answer.

7 0

2 years ago

(Write the equation for the line that goes through -3,-3) slope 7/3

grin007 [14]

$\bf (\stackrel{x_1}{-3}~,~\stackrel{y_1}{-3})~\hspace{10em} \stackrel{slope}{m}\implies \cfrac{7}{3} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-3)}=\stackrel{m}{\cfrac{7}{3}}[x-\stackrel{x_1}{(-3)}] \implies y+3=\cfrac{7}{3}(x+3) \\\\\\ y+3=\cfrac{7}{3}x+7\implies y=\cfrac{7}{3}x+4$

3 0

2 years ago