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Neko [114]
3 years ago
14

Which ordered pair is also on the line relating the area of a face of a cube and the surface area of the cube?

Mathematics
1 answer:
Alexxx [7]3 years ago
7 0
What cube how big is the cube
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A spinner has 3 sections of different sizes. The sections are orange, purple, and green. 1/4 of the spinner is orange. 7/12 of t
Harman [31]

Answer: it would be 2/3

Step-by-step explanation:

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Which of the following are true statement about the slope of the line
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Step-by-step explanation:

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3 years ago
What's the sum of an infinite geometric series if the first term is 156 and the common ratio is 2∕3?
DENIUS [597]

Answer:

468

Step-by-step explanation:

Formula for infinite sum of geometric series is;

S_∞ = a1/(1 - r)

Where;

a1 is first term

r is common ratio

We are given;

a1 = 156

r = ⅔

Thus;

S_∞ = 156/(1 - ⅔)

S_∞ = 156/(⅓)

S_∞ = 468

4 0
3 years ago
PLEASE HELP!! QUIZ TOMORROW ☹️
Sunny_sXe [5.5K]
The right answer is c

6 0
3 years ago
Mathematical induction, prove the following two statements are true
adelina 88 [10]
Prove:
1+2\left(\frac12\right)+3\left(\frac12\right)^{2}+...+n\left(\frac12\right)^{n-1}=4-\dfrac{n+2}{2^{n-1}}
____________________________________________

Base Step: For n=1:
n\left(\frac12\right)^{n-1}=1\left(\frac12\right)^{0}=1
and
4-\dfrac{n+2}{2^{n-1}}=4-3=1
--------------------------------------------------------------------------

Induction Hypothesis: Assume true for n=k. Meaning:
1+2\left(\frac12\right)+3\left(\frac12\right)^{2}+...+k\left(\frac12\right)^{k-1}=4-\dfrac{k+2}{2^{k-1}}
assumed to be true.

--------------------------------------------------------------------------

Induction Step: For n=k+1:
1+2\left(\frac12\right)+3\left(\frac12\right)^{2}+...+k\left(\frac12\right)^{k-1}+(k+1)\left(\frac12\right)^{k}

by our Induction Hypothesis, we can replace every term in this summation (except the last term) with the right hand side of our assumption.
=4-\dfrac{k+2}{2^{k-1}}+(k+1)\left(\frac12\right)^{k}

From here, think about what you are trying to end up with.
For n=k+1, we WANT the formula to look like this:
1+2\left(\frac12\right)+...+k\left(\frac12\right)^{k-1}+(k+1)\left(\frac12\right)^{k}=4-\dfrac{(k+1)+2}{2^{(k+1)-1}}

That thing on the right hand side is what we're trying to end up with. So we need to do some clever Algebra.

Combine the (k+1) and 1/2, put the 2 in the bottom,
=4-\dfrac{k+2}{2^{k-1}}+\dfrac{(k+1)}{2^{k}}

We want to end up with a 2^k as our final denominator, so our middle term is missing a power of 2. Let's multiply top and bottom by 2,
=4+\dfrac{-2(k+2)}{2^{k}}+\dfrac{(k+1)}{2^{k}}

Distribute the -2 and combine the fractions together,
=4+\dfrac{-2k-4+(k+1)}{2^{k}}

Combine like-terms,
=4+\dfrac{-k-3}{2^{k}}

pull the negative back out,
=4-\dfrac{k+3}{2^{k}}

And ta-da! We've done it!
We can break apart the +3 into +1 and +2,
and the +0 in the bottom can be written as -1 and +1,
=4-\dfrac{(k+1)+2}{2^{(k-1)+1}}
3 0
3 years ago
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