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2 years ago

5

HELPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPPP PLEASEMPLEASEPLEASPLEPLDAPDKSOFJWDIEUNDW BEGGING 0PLEASE PLEASE PLEASE PLAS PQLDWPPLE

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2 answers:

nordsb [41]2 years ago

8 0

Answer: 21

Step-by-step explanation: Just plug in the numbers for x, and you get 21 when you plug that in.

tiny-mole [99]2 years ago

6 0

Answer:

D 9

Step-by-step explanation:

hmu if you need more help! :)

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A bouncing ball reaches a height of 54 inches at its first peak, 36 inches at its second peak, and 24 inches at its third peak.

steposvetlana [31]

54, 36, 24 are the 1st 3 element of a geometric progression with 2/3 as a common ratio: PROOF:
the 1st term is 54, (a₁= 54) the 2nd term a₂ = 24, then
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So 2/3 is common ratio. We know that :U(n) = a.(r)ⁿ⁻¹. Then if a =54 and r = x (given by the problem), then f(x) = 54.xⁿ⁻¹
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2 years ago

Read 2 more answers

Can you guys help me out on this? I'm still learning sign, cosign, and tangent :)

Yakvenalex [24]

Answer:

$\sin d = \frac{4}{7}$ ; $\sin e = \frac{\sqrt{33} }{7}$

$\cos d = \frac{\sqrt{33} }{7}$ ; $\cos e = \frac{4}{7}$

$\tan d = \frac{4}{\sqrt{33} }$ ; $\tan e = \frac{\sqrt{33} }{4}$

Step-by-step explanation:

For a right angled triangle with one of its angle α (alpha) :-

$\sin \alpha = \frac{Side \: opposite \: to \: \alpha }{Hypotenuse \: of \: the \: triangle}$

$\cos \alpha = \frac{Side \: adjacent \: to \: \alpha }{Hypotenuse \: of \: the \: triangle}$

$\tan \alpha = \frac{Side \: opposite \: to \: \alpha }{Side \: adjacent \: to \: \alpha }$

__________________________________________________

According to the question ,

1) When α (alpha) = d

$\sin d = \frac{4}{7}$

$\cos d = \frac{\sqrt{33} }{7}$

$\tan d = \frac{4}{\sqrt{33} }$

2) When α (alpha) = e

$\sin e = \frac{\sqrt{33} }{7}$

$\cos e = \frac{4}{7}$

$\tan e = \frac{\sqrt{33} }{4}$

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Step-by-step explanation:

I was only able to finish one through eight The rest follow the same concept though

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