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snow_lady [41]
3 years ago
8

Find the parial sum of the first 8 terms for the series 8 +16 +32+

Mathematics
1 answer:
Ira Lisetskai [31]3 years ago
6 0
2040   8,16,32,64,128,256,512,1024
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Find the answer given below​
Anastasy [175]

Answer:

Surface area = 6l^2

Volume = l^3

If total surface area increased by 2, then the length increased by \sqrt{2}

Volume then is increased by (\sqrt{2}l)^3 = 2\sqrt{2}l^3

1 :2 \sqrt{2}

Which is not an option in the question so this question clearly has a problem.

First of all It's supposed to be "then" not "than," so I'm not sure who teaches math without even knowing this basic grammar.

Second of all, if the answer is 8:1 the question should be "If <u>each sides</u> of a cube is doubled" not "the lateral surface area." Come to think of it what even is lateral surface area of a cube if all sides of the cube is supposed to be same.

Or, if the answer is 2:1 then, it should be If height lateral surface area of a cube is doubled <u>by doubling the height and without changing the length and width</u>"  But, then the the shape would no longer be a cube.

Who even wrote this garbage question what.

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2 years ago
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kirill [66]

Answer:

21

Step-by-step explanation:

Add all the numbers together

6+8+2.5+2.5+2=21

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3 years ago
What angle relationship describes VQR and QRW?
Vlad [161]

Answer: alternate interior angles

Step-by-step explanation:

Got it Right on the test. :)

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PLEASE HELP! I'll give 5 out of 5 stars, give thanks, and give as many points as I can.
Liula [17]

Answer:

\boxed{\boxed{x=\dfrac{\pi}{3}\ \vee\ x=\pi\ \vee\ x=\dfrac{5\pi}{3}}}

Step-by-step explanation:

\cos(3x)=-1\iff3x=\pi+2k\pi\qquad k\in\mathbb{Z}\\\\\text{divide both sides by 3}\\\\x=\dfrac{\pi}{3}+\dfrac{2k\pi}{3}\\\\x\in[0,\ 2\pi)

\text{for}\ k=0\to x=\dfrac{\pi}{3}+\dfrac{2(0)\pi}{3}=\dfrac{\pi}{3}+0=\boxed{\dfrac{\pi}{3}}\in[0,\ 2\pi)\\\\\text{for}\ k=1\to x=\dfrac{\pi}{3}+\dfrac{2(1)\pi}{3}=\dfrac{\pi}{3}+\dfrac{2\pi}{3}=\dfrac{3\pi}{3}=\boxed{\pi}\in[0,\ 2\pi)\\\\\text{for}\ k=2\to x=\dfrac{\pi}{3}+\dfrac{2(2)\pi}{3}=\dfrac{\pi}{3}+\dfrac{4\pi}{3}=\boxed{\dfrac{5\pi}{3}}\in[0,\ 2\pi)\\\\\text{for}\ k=3\to x=\dfrac{\pi}{3}+\dfrac{2(3)\pi}{3}=\dfrac{\pi}{3}+\dfrac{6\pi}{3}=\dfrac{7\pi}{3}\notin[0,\ 2\po)

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3 years ago
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Answer:

45x happy to help ya :)

Step-by-step explanation:

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