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Bogdan [553]
3 years ago
6

I need answers for 13 and 14 please

Mathematics
2 answers:
allsm [11]3 years ago
8 0

180-72=108

108/3=x

x=36

B=36x2=72

Nataly_w [17]3 years ago
3 0

Answer:

(13) Equation: x + (x+2) + 72 = 180

      Answer: x = 53

(14) B = 55 degrees

Step-by-step explanation:

(13) The sum of the angles in any triangle is always 180 degrees, so you can find the sum of the given angles and set up an equation, then solve

x + (x+2) + 72 = 180

2x + 74 = 180

2x = 106

x = 53

(14) If x = 53 and angle B is equal to to x + 2, you can just add 2 to 53 and get the answer that the measure of angle B is 55 degrees

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Step-by-step explanation:

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2 years ago
A triangle has an area of 56 square units it's height is 14 units
kobusy [5.1K]
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3 years ago
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:

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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
Help!! Which of the following best represents Z1 • Z2 select all that apply
AURORKA [14]

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\sin \left(\frac{\pi }{4}\right)=\frac{\sqrt{2}}{2}

\cos \left(\frac{3\pi }{4}\right)=-\frac{\sqrt{2}}{2}

\sin \left(\frac{3\pi }{4}\right)=\frac{\sqrt{2}}{2}\\

Z_1*Z_2=\sqrt{3}\left(cos\:\frac{\pi }{4}+i\:sin\:\frac{\pi }{4}\right)\cdot \sqrt{6}\left(cos\:\frac{3\pi \:}{4}+i\:sin\:\frac{3\pi \:}{4}\right)

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On simplifying, we get

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<h2>Therefore, correct option is  1st option.</h2>
3 0
3 years ago
how many such tests would it take for the probability of committing at least one type i error to be at least 0.7? (round your an
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The number of tests that it would take for the probability of committing at least one type I error to be at least 0.7 is 118   .

In the question ,

it is given that ,

the probability of committing at least , type I error is = 0.7

we  have to find the number of tests ,

let the number of test be n ,

the above mentioned situation can be written as

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which is written as ,

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On further simplification ,

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Therefore , the number of tests are 118 .

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brainly.com/question/17062640

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