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schepotkina [342]
4 years ago
8

Simplify the expression: 4(2x-7) * 6x-7 08x-7 0 2x-28 8x-28

Mathematics
2 answers:
ankoles [38]4 years ago
7 0

<em>So</em><em> </em><em>the</em><em> </em><em>right</em><em> </em><em>answer</em><em> </em><em>is</em><em> </em><em>of</em><em> </em><em>option</em><em> </em><em>D</em><em>.</em>

<em>Look</em><em> </em><em>at</em><em> </em><em>the</em><em> attached</em><em> </em><em>picture</em>

<em>H</em><em>ope</em><em> </em><em>it</em><em> </em><em>will</em><em> </em><em>help</em><em> </em><em>you</em><em>.</em><em>.</em><em>.</em>

<em>G</em><em>ood</em><em> </em><em>luck</em><em> </em><em>on</em><em> </em><em>your</em><em> </em><em>assignment</em>

<em>~</em><em>p</em><em>r</em><em>a</em><em>g</em><em>y</em><em>a</em>

oksian1 [2.3K]4 years ago
3 0

Answer:

Answer is (d)

Step-by-step explanation:

4(2x - 7)

8x - 28

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I NEED HELP PLEASE! :)<br> thanks!
torisob [31]

Since ΔABC ~ ΔEDC, ∠B = ∠D.

Since both triangles appear to be similar, the corresponding angles are the same, and corresponding sides are the same or have the same ratio.

We can write an equation to resemble the problem:

8x + 16 = 120

Solve for x.

8x + 16 = 120

~Subtract 16 to both sides

8x + 16 - 16 = 120 - 16

~Simplify

8x = 104

~Divide 8 to both sides

8x/8 = 104/8

~Simplify

x = 13

Therefore, the answer is 13.

Best of Luck!

6 0
3 years ago
Please include step by step!
Paladinen [302]

Answer:

C=10pi

Step-by-step explanation:

5 0
2 years ago
Read 2 more answers
WILL MARK BRAINLIEST
Ivanshal [37]

Answer:

27

Step-by-step explanation:

First you need to find x.

It is shown that DE=EF, so we can use this.

5x-3=3x+7

5x=3x+10

2x=10

x=5

Then, you plug in the value of the variable and solve.

6(5)-3

30-3

27 = DF

8 0
3 years ago
Read 2 more answers
Look at the image. (calculus)
mash [69]
<h3>Answer: Choice H)  2</h3>

=============================================

Explanation:

Recall that the pythagorean trig identity is \sin^2 x + \cos^2x = 1

If we were to isolate sine, then,

\sin^2 x + \cos^2x = 1\\\\\sin^2 x = 1-\cos^2x\\\\\sin x = \sqrt{1-\cos^2x}\\\\

We don't have to worry about the plus minus because sine is positive when 0 < x < pi/2.

Through similar calculations, \cos x = \sqrt{1-\sin^2x}\\\\

Cosine is also positive in this quadrant.

-------------

So,

\frac{\sqrt{1-\cos^2x}}{\sin x}+\frac{\sqrt{1-\sin^2x}}{\cos x}\\\\\frac{\sin x}{\sin x}+\frac{\cos x}{\cos x}\\\\1+1\\\\2

Therefore,

\frac{\sqrt{1-\cos^2x}}{\sin x}+\frac{\sqrt{1-\sin^2x}}{\cos x}=2

is an identity as long as 0 < x < pi/2

5 0
2 years ago
Read 2 more answers
30,135x0.15 hgggggggggggggggggggggggggggggggggggggggggggggggggggggggggnnnn tyyyyyyyy
elena-s [515]

Answer:

what

Step-by-step explanation:

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7 0
3 years ago
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