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LUCKY_DIMON [66]
2 years ago
6

I need help pls and thx

Mathematics
2 answers:
padilas [110]2 years ago
7 0

Answer:

1 - Quadrant I

2 - (5, -3), Quadrant IV

3 - (-5, -3), Quadrant III

Step-by-step explanation:

Assoli18 [71]2 years ago
3 0

Answer:

The image isnt loading is there any way you can put it in agian?

Step-by-step explanation:

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3 years ago
Given the following right triangle, find cose, sine, tane sececsce, and cote. Do not
IgorLugansk [536]

Step-by-step explanation:

Use the Pythagorean theorem:

leg^2+leg^2=hypotenuse^2

We have

leg=4,\ hypotenuse=10

Substitute:

4^2+leg^2=10^2

16+leg^2=100             <em>subtract 16 from both sides</em>

leg^2=84\to leg=\sqrt{84}\\\\leg=\sqrt{(4)(21)}\\\\leg=\sqrt4\cdot\sqrt{21}\\\\leg=2\sqrt{21}

sine=\dfrac{opposite}{hypotenuse}\\\\cosine=\dfrac{adjacent}{hypotenuse}\\\\tangent=\dfrac{opposite}{adjacent}\\\\cotangent=\dfrac{adjacent}{opposite}\\\\secant=\dfrac{hypotenuse}{adjacent}\\\\cosecant=\dfrac{hypotenuse}{opposite}

Substitute:

hypotenuse=10,\ opposite=4,\ adjacent=2\sqrt{21}

\sin\theta=\dfrac{4}{10}=\dfrac{2}{5}\\\\\cos\theta=\dfrac{2\sqrt{21}}{10}=\dfrac{\sqrt{21}}{5}\\\\\tan\theta=\dfrac{4}{2\sqrt{21}}=\dfrac{2}{\sqrt{21}}\cdot\dfrac{\sqrt{21}}{\sqrt{21}}=\dfrac{2\sqrt{21}}{21}\\\\\cot\theta=\dfrac{2\sqrt{21}}{4}=\dfrac{\sqrt{21}}{2}\\\\\sec\theta=\dfrac{10}{2\sqrt{21}}=\dfrac{5}{\sqrt{21}}\cdot\dfrac{\sqrt{21}}{\sqrt{21}}=\dfrac{5\sqrt{21}}{21}\\\\\csc\theta=\dfrac{10}{4}=\dfrac{5}{2}

5 0
3 years ago
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