2 years ago

PLEASEEE HELP BIG FINALS TEST !!!! :( plssssss

Mathematics

1 answer:

Sati [7]2 years ago

7 0

I’m pretty sure I know the answers but could you tell me the options for the drop boxes:)

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Snowcat [4.5K]

The yellow one. the angles will always be the same no matter how small

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2 years ago

Is the following true for all positive values of a and b: If a^2>b^2, then a>b

Dominik [7]

Answer:

Yes, that is correct.

Step-by-step explanation:

a would always have to be greater than b since you're just multiplying each variable by itself.

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3 years ago

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Nina thought of a number. She called her number y. She added 14 to her number, divided the result by 6, and she got 12. What was

Shalnov [3]

Y=58

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72-14=58

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3 years ago

Which reason justifies the last step in a proof that ΔEDG≌ΔADC?

Alja [10]

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3 years ago

In this exercise, consider a particle moving on a circular path of radius b described by r(t) = b cos(ωt)i + b sin(ωt)j, where ω

Bingel [31]

Answer:

Acceleration of the particle = $bw^{2}$

Step-by-step explanation:

We are given the position vector of a particle moving in a circle of radius b units.

r(t) = b cos(ωt)i + b sin(ωt)j

Velocity , v = $\frac{dr}{dt}$ = -bω sin(ωt)i + bω cos(ωt)j

The magnitude of velocity, v = $\sqrt{v_x^{2} +v_y^{2} }$

Squaring both sides,

$v^{2} = b^{2} w^{2}(sin^{2}(wt)+cos^{2}(wt))$

Since $sin^{2}(wt)+cos^{2}(wt))$ = 1

$v^{2} = b^{2}w^{2}$

The acceleration towards the centre is called the centripetal acceleration and is given by

a = $\frac{v^{2} }{r}$

a = $\frac{b^{2}w^{2}}{b}$

a = $bw^{2}$

3 0

2 years ago