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3 years ago

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Please help me I need help

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vivado [14]3 years ago

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Figure ABCD - Figure EFGH

Colt1911 [192]

<h3><u>Answer:</u></h3>

30.5 units

<h3><u>Step-by-step explanation:</u></h3>

We know that:

<u>Trapezoid ABCD has one side that is 25 unit. </u>
<u>The other 3 sides (AB, BC, and CD) are 12 units. </u>
<u>EF = 6 units</u>
<u>EH = 12.5</u>
<u>Trapezoid ABCD is congruent to Trapezoid EFGH</u>

<em>If the 3 sides of the trapezoid ABCD are the same sides, then side EF, FG, GH must be of the same length because of congruence. The value of FG and GH must be the same length as EF. We can clearly see in the picture that EF is 6 units. Hence, EF is 6 units, FG is 6 units, and GH is 6 units. The work of the perimeter is shown below.</em>

<u>Work</u>

=> 6(3) + 12.5
=> 18 + 12.5
=> <u>30.5 units</u>

Hence, the perimeter of EFGH is 30.5 units.

$BrainiacUser1357$

4 0

2 years ago

Good afternoon, can you help me please

Alexus [3.1K]

It would just be 14-10 each week so 14-10 14-10 14-10 and so on

8 0

3 years ago

Write the next two numbers in the sequence by looking at patterns 3, 9 ,27 ,81, _, _

kumpel [21]

3, 9, 27, 81, 243, 729

5 0

3 years ago

Read 2 more answers

If you were to deposit $1,000 into an account that paid 10 percent interest compounded semiannually, how much money would you ha

dexar [7]

Semiannual means there will be two total periods in a year. Divide ten percent by two to get the interest rate per period. Then multiple by 1+i twice

6 0

2 years ago

I need help with this question

Novay_Z [31]

Answer:

$\frac{\sqrt{3} - 1}{2\sqrt{2}}$

$\frac{-(\sqrt{3} + 1)}{2\sqrt{2}}$

$- \frac{\sqrt{3} - 1}{\sqrt{3} + 1}$

Step-by-step explanation:

Given $\frac{11 \pi}{12} = \frac{3 \pi}{4} + \frac{\pi}{6}$

(A) $sin(\frac{11\pi}{12}) = sin (\frac{3 \pi}{4} + \frac{\pi}{6})$

We know that Sin(A + B) = SinA cosB + cosAsinB

Substituting in the above formula we get:

$sin (\frac{3\pi}{4} + \frac{\pi}{6}) = \frac{1}{\sqrt{2}} . \frac{\sqrt{3}}{2} + \frac{-1}{\sqrt{2}}. \frac{1}{2}$

$$ \implies \frac{1}{\sqrt{2}} (\frac{\sqrt{3} - 1}{2}) = \frac{\sqrt{3} - 1}{2\sqrt{2}}$

(B) Cos(A + B) = CosAcosB - SinASinB

$cos(\frac{11\pi}{12}) = cos(\frac{3\pi}{4} + \frac{\pi}{6}})$

$\implies \frac{-1}{\sqrt{2}}. \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}} . \frac{1}{2}$

$\implies cos(\frac{11\pi}{12}) = cos(\frac{3\pi}{4} + \frac{\pi}{6})$

$$ = \frac{-(\sqrt{3} + 1)}{2\sqrt{2}}$

(C) Tan(A + B) = $\frac{Sin(A +B)}{Cos(A + B)}$

From the above obtained values this can be calculated and the value is $- \frac{\sqrt{3} - 1}{\sqrt{3} + 1}$ .

3 0

3 years ago