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3 years ago

Can you help me finish this paper?? I need the answers ASAP!! Plz I will give you 79 points

Mathematics

2 answers:

BlackZzzverrR [31]3 years ago

7 0

ok so you add ok 4. is 23 4 plus 6 plus 8 plus 5 is 23 and you repeat and add on every the first 1. is 16

Yuki888 [10]3 years ago

4 0

Answer:

Step-by-step explanation:

1）22

2）40

3）44

4）62

5）26

6）55

7）29

8）45

9)The picture is not clear

10) 52

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Sholpan [36]

Given:

$\cos \theta =\dfrac{3}{5}$

$\sin \theta$

To find:

The quadrant of the terminal side of $\theta$ and find the value of $\sin\theta$ .

Solution:

We know that,

In Quadrant I, all trigonometric ratios are positive.

In Quadrant II: Only sin and cosec are positive.

In Quadrant III: Only tan and cot are positive.

In Quadrant IV: Only cos and sec are positive.

It is given that,

$\cos \theta =\dfrac{3}{5}$

$\sin \theta$

Here cos is positive and sine is negative. So, $\theta$ must be lies in Quadrant IV.

We know that,

$\sin^2\theta +\cos^2\theta =1$

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$\sin \theta=\pm \sqrt{1-\cos^2\theta}$

It is only negative because $\theta$ lies in Quadrant IV. So,

$\sin \theta=-\sqrt{1-\cos^2\theta}$

After substituting $\cos \theta =\dfrac{3}{5}$ , we get

$\sin \theta=-\sqrt{1-(\dfrac{3}{5})^2}$

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MakcuM [25]

Answer:

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Step-by-step explanation:

You must find the cube root of 15.

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hope it helps! :-)

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