Task E

Mathematics

1 answer:

Anvisha [2.4K]2 years ago

3 0

Answer:

39.36

Step-by-step explanation:

We multiply 9.84 by 4 which gives us the answer above

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Type the correct answer for each.

xxTIMURxx [149]

If point $\left(x,\ \frac{\sqrt{7}}{3}\right)$ is on the unit ircle, then:

$x^2+\left( \frac{ \sqrt{7} }{3} \right)^2=1 \\ \\ \Rightarrow x^2+ \frac{7}{9} =1 \\ \\ \Rightarrow x^2=1- \frac{7}{9} = \frac{2}{9} \\ \\ \Rightarrow x= \sqrt{ \frac{2}{9} } = \frac{ \sqrt{2} }{3}$

Since, the point is in the second quadrant, x is negative.

Thus, $(x,\ y)=\left(x,\ \frac{\sqrt{7}}{3}\right)=\left(-\frac{ \sqrt{2} }{3},\ \frac{\sqrt{7}}{3}\right)$

Part A:

$3 \sqrt{7} =6\left( \frac{ \sqrt{7} }{2} \right) \\ \\ =6\left( \frac{ \sqrt{7} }{3} \cdot \frac{3}{ \sqrt{2} } \right) \\ \\ =6\left( \frac{ \frac{ \sqrt{7} }{3} }{ \frac{ \sqrt{2} }{3} } \right)=-6\left( \frac{ \frac{ \sqrt{7} }{3} }{ -\frac{ \sqrt{2} }{3} } \right) \\ \\ =-6\left( \frac{y}{x} \right)=-6\tan\theta$

Therefore, $-6\tan\theta=3\sqrt{7}$ .

Part B:

$- \frac{ \sqrt{7} }{2} =- \frac{ \sqrt{7} }{3} \cdot \frac{3}{ \sqrt{2} } \\ \\ =- \frac{ \frac{ \sqrt{7} }{3} }{ \frac{ \sqrt{2} }{3} } = \frac{ \frac{ \sqrt{7} }{3} }{ -\frac{ \sqrt{2} }{3} } = \frac{y}{x} \\ \\ =\tan\theta$

Therefore, $\tan\theta=- \frac{ \sqrt{7} }{2}$

8 0

3 years ago

What the value of p in the question shown

hammer [34]

It would be D because you have to do the cube root on both sides. What I do is ignore the decimal and add it in later once the math is complete like on this question you can tell by process of elimination that 8 squared is 64 but you’re look for the cube root so that eliminates a and c and now just make sure the decimal is in the right place

5 0

3 years ago

Giving away points! <br> =D

Rzqust [24]

Answer:

Thanks you very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very very much for points

6 0

3 years ago

Read 2 more answers

Would each of them be able to be the side lengths of a triangle?

ratelena [41]

Pythagoras Theorem:
hipotenuse²=leg₁²+leg₂²

First posible triangle:
hypotenuse=13 (13²=169)
leg₁=12 ( 12²=144)
leg₂=5 (5²=25)

13³=144 + 25

Answer:can be side lengths of a triangle

Second triangle:
hypotenuse=12.6 (12.6²=158.76)
leg₁=6.7 ( 6.7²=44.89)
leg₂=6.5 (6.5²=42.25)

leg₁²+leg₂²=44.89+42.25=87.14≠158.76

Answer: cannot be side lenghts of a triangle.

third triangle:
hypotenuse=13 (13²=169)
leg₁=12 ( 12²=144)
leg₂=11 (11²=121)

leg₁²+leg₂²=144+121=265≠169

Answer: cannot be side lenghts of a triangle.

fourth triangle:
hypotenuse=13 (13²=169)
leg₁=6 ( 6²=36)
leg₂=4 (4²=16)

leg₁²+leg₁²=36+16=52≠169

Answer: cannot be side lenghts of a triangle.

8 0

3 years ago

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Show one way to count $82 to $512

mylen [45]

What do u mean like 82+82?

4 0

3 years ago

Read 2 more answers