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Svetradugi [14.3K]
2 years ago
11

Help please!!! i dont get it

Mathematics
1 answer:
IgorLugansk [536]2 years ago
5 0

The amount of money in the account if compounded quarterly at 6% after 9 years is $22,217

<h3>Compound interest</h3>

  • Principal, P = $13,000
  • Time, t = 9 years
  • Interest rate, r = 6% = 0.06
  • Number of periods, n = 4

A = P(1 + r/n)^nt

= 13,000(1 + 0.06/4)^4×9

= 13,000(1 + 0.015)^36

= 13,000(1.015)^36

= 13000(1.709)

= $22,217

Learn more about compound interest:

brainly.com/question/24924853

#SPJ1

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Gwar [14]
Line a and Line b are parallel.

∠x = 34° (alternative angles on a parallel line are equal)

Answer: ∠x = 34°  (option C)
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3 years ago
20 = 10x - 6(2x + 5)
viva [34]

Answer:

-25=x

Step-by-step explanation:

20=10x-6(2x+5)

20=10x-12x-30

20=-2x-30

+30   +30

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50=-2x

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-25=x

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3 years ago
Please help me answer Question 1, and 3! Thanks!!!
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I hope this helps you

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3 years ago
Explain why Rolle's Theorem does not apply to the function even though there exist a and b such that f(a) = f(b).
aliya0001 [1]
This may help you

he functión Cot[x/2] is not continuos in the points <span>x=2nπ,  where  n=0,1,2,3,...</span> You can check it with a calculator. So the function is not continuos in the domain the problem gives you, so the Rolle's theorem can not be applied. If the inteverval was, <span>[π/2,3π/2]</span> Then we could apply the Rolle's theorem.

5 0
4 years ago
If cos Θ = square root 2 over 2 and 3 pi over 2 &lt; Θ &lt; 2π, what are the values of sin Θ and tan Θ? sin Θ = square root 2 ov
densk [106]

Answer:

\huge\boxed{\sin\theta=-\dfrac{\sqrt2}{2};\ \tan\theta=-1}

Step-by-step explanation:

We have:

\\cos\theta=\dfrac{\sqrt2}{2},\ \dfrac{3\pi}{2}

For sine use:

\sin^2x+\cos^2x=1\to\sin^2x=1-\cos^2x

Substitute:

\sin^2\theta=1-\left(\dfrac{\sqrt2}{2}\right)^2\\\\\sin^2\theta=1-\dfrac{(\sqrt2)^2}{2^2}\\\\\sin^2\theta=1-\dfrac{2}{4}\\\\\sin^2\theta=\dfrac{4}{4}-\dfrac{2}{4}\\\\\sin^2\theta=\dfrac{4-2}{4}\\\\\sin^2\theta=\dfrac{2}{4}\to\sin\theta=\pm\sqrt{\dfrac{2}{4}}\\\\\sin\theta=\pm\dfrac{\sqrt2}{\sqrt4}\\\\\sin\theta=\pm\dfrac{\sqrt2}{2}

θ in IV quadrant, therefore sine is negative.

\sin\theta=-\dfrac{\sqrt2}{2}

For tangent use:

\tan x=\dfrac{\sin x}{\cos x}

Substitute:

\tan\theta=\dfrac{-\frac{\sqrt2}{2}}{\frac{\sqrt2}{2}}=-\dfrac{\sqrt2}{2}\cdot\dfrac{2}{\sqrt2}=-1

8 0
3 years ago
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