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Tems11 [23]
3 years ago
12

(

ula1" title="( {8x}^{2} + 4x + 4) - (7x + 9)" alt="( {8x}^{2} + 4x + 4) - (7x + 9)" align="absmiddle" class="latex-formula">
solve​
Mathematics
1 answer:
son4ous [18]3 years ago
4 0

Answer:

8x^{2} -3x-5

Step-by-step explanation:

8x^{2} +4x+4-7x-9

8x^{2} +4x-7x+4-9

8x^{2} -3x-5

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Triangle UVW is formed by the three squares G, H, and I:
Alinara [238K]

Answer:

The correct answer would be C- (UV)2 + (VW)2 = (UW)2, because 81 + 144 = 225

-Thank you

3 0
3 years ago
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Prove that (Root of Sec A - 1 / Root of Sec A + 1) + (Root of Sec A + 1 / Root of Sec A - 1) = 2 cosec A
iVinArrow [24]

Answer:

The answer is below

Step-by-step explanation:

We need to prove that:

(Root of Sec A - 1 / Root of Sec A + 1) + (Root of Sec A + 1 / Root of Sec A - 1) = 2 cosec A.

Firstly, 1 / cos A = sec A, 1 / sin A = cosec A and tanA = sinA / cosA.

Also, 1 + tan²A = sec²A; sec²A - 1 = tan²A

\frac{\sqrt{secA-1} }{\sqrt{secA+1} } +\frac{\sqrt{secA+1} }{\sqrt{secA-1} } =\frac{(\sqrt{secA-1)}(\sqrt{secA-1})+(\sqrt{secA+1)}(\sqrt{secA+1}) }{(\sqrt{secA+1})(\sqrt{secA-1}) } \\\\=\frac{secA-1+(secA+1)}{\sqrt{sec^2A-secA+secA-1} } \\\\=\frac{2secA}{\sqrt{sec^2A-1} } \\\\=\frac{2secA}{\sqrt{tan^2A} } \\\\=\frac{2secA}{tanA} \\\\=\frac{2*\frac{1}{cosA} }{\frac{sinA}{cosA} }\\\\= 2*\frac{1}{cosA}*\frac{cosA}{sinA}\\\\=2*\frac{1}{sinAA}\\\\=2cosecA

7 0
3 years ago
I HAVE AN EXAM PLEASE HELP ME GET THE ANSWER
sesenic [268]
Answer: The ladder is 5.3 m long
Step by step solution
1. Draw out the scenario, ie a right triangle where the hypotenuse is the ladder, the wall is 4.5m and the angle the ladder and the ground makes 58 degree.

2. Soh, Cah, Toa
The wall is opposite, the ground is adjacent, and the ladder is hypotenuse
Sine (58) = 4.5/h

3. Solve for hypotenuse
h * sine (58) = 4.5
h = 4.5/sine (58)
h = ~ 5.3 m
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Answer:

ok put me brainliest  please

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Step-by-step explanation:

6 0
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Find the coordinates of the vertices of each figure after the given translation. Then graph the translation image.
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