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ki77a [65]
3 years ago
14

8+3b+-13 WHO WANTS 99 POINTS XD

Mathematics
2 answers:
BARSIC [14]3 years ago
5 0

Answer:

\Huge\boxed{3b-5}

Step-by-step explanation:

\Large\boxed{\mathsf{SUBJECT: MATH}}}

Isolate b on one side of the equation.

First, subtract.

8-13=-5

3b-5

\Large\boxed{\mathsf{\Rightarrow 3b-5}}

So, the correct answer is 3b-5.

LenKa [72]3 years ago
4 0

Answer:

3b - 5

Step-by-step explanation:

8 + 3b + - 13

Combine like terms

8 +-13 = 8 - 13

8 - 13 = -5

Write it out together

3b - 5

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Check the picture below.

\bf \textit{using the pythagorean theorem}
\\\\
c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b
\qquad 
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\end{cases}
\\\\\\
\sqrt{41^2-(-9)^2}=b\implies \sqrt{1681-81}=b\\\\\\ \sqrt{1600}=b\implies 40=b\\\\
-------------------------------

\bf sin(\theta )=\cfrac{\stackrel{opposite}{40}}{\stackrel{hypotenuse}{41}}\qquad~~  cos(\theta )=\cfrac{\stackrel{adjacent}{-9}}{\stackrel{hypotenuse}{41}}\qquad~~  tan(\theta )=\cfrac{\stackrel{opposite}{40}}{\stackrel{adjacent}{-9}}
\\\\\\
csc(\theta )=\cfrac{\stackrel{hypotenuse}{41}}{\stackrel{opposite}{40}}\qquad ~~sec(\theta )=\cfrac{\stackrel{hypotenuse}{41}}{\stackrel{adjacent}{-9}}\qquad ~~cot(\theta )=\cfrac{\stackrel{adjacent}{-9}}{\stackrel{opposite}{40}}

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3 years ago
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