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Papessa [141]
3 years ago
13

Hurry i'll give brainiest if u help me

Mathematics
1 answer:
galben [10]3 years ago
6 0
X^2 = b^2-c^2
x^2= sqrt117^2-6^2
x^2=sqrt117×sqrt117-36
x^2=117-36
x^2=81
x=sqrt81
x=9cm
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26 = 2(3x - 2)<br> please help
Nadya [2.5K]

Answer:x=5

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Step-by-step explanation:

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I need help with Geometry
ehidna [41]

Answer:

We conclude that segment QR is the shortest.

Hence, option B is true.

Step-by-step explanation:

First, we need to determine the missing angle m∠R

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m∠Q = 83°

m∠R = ?

We know the sum of angles of a triangle is 180°.

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Therefore, we conclude that segment QR is the shortest.

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3 years ago
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Check the picture below.

\bf \textit{using the pythagorean theorem}&#10;\\\\&#10;c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b&#10;\qquad &#10;\begin{cases}&#10;c=hypotenuse\\&#10;a=adjacent\\&#10;b=opposite\\&#10;\end{cases}&#10;\\\\\\&#10;\sqrt{41^2-(-9)^2}=b\implies \sqrt{1681-81}=b\\\\\\ \sqrt{1600}=b\implies 40=b\\\\&#10;-------------------------------

\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}}&#10;\\\\\\&#10;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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Last questions on this quiz (I always get confused on finding the x value)
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