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lutik1710 [3]
3 years ago
15

(5-2i)+(3+4i)-(-6+2i)

Mathematics
1 answer:
pav-90 [236]3 years ago
3 0

Simplify

5 - 2i + 3 + 4i - (-6 + 2i)

Simplify brackets

5 - 2i + 3 + 4i + 6 - 2i

Collect like terms

(5 + 3 + 6) + (-2i + 4i - 2i)

Simplify

<u>= 14</u>

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The distance from the base of the ramp to the base of the sofa rounded to the nearest tenth is 47.3 inches.

<u><em>Explanation</em></u>

According to the diagram, length of the ramp is, AC= 55 inches. The top of the seat cushion is 28 inches above the floor. That means, BC=28 inches

We need to find the distance from the base of the ramp to the base of the sofa which is AB

In right angle triangle ABC, using Pythagorean theorem...

AC^2 = AB^2 + BC^2\\ \\ (55)^2= AB^2 + (28)^2 \\ \\ AB^2 = (55)^2 - (28)^2 \\ \\ AB^2 = 3025 - 784\\ \\ AB^2 = 2241\\ \\ AB = \sqrt{2241}= 47.3

So, the distance from the base of the ramp to the base of the sofa rounded to the nearest tenth is 47.3 inches.

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The sum of 30 times (1/3)^(n-1) from 1 to infinity
sergeinik [125]

Let

S_n=\displaystyle1+\frac13+\frac1{3^2}+\cdots+\frac1{3^n}

Then

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and

S_n-\dfrac13S_n=\dfrac23S_n=1-\dfrac1{3^{n+1}}\implies S_n=\dfrac32-\dfrac1{2\cdot3^n}

and as n\to\infty, we end up with

\displaystyle\lim_{n\to\infty}S_n=\lim_{n\to\infty}\sum_{i=1}^{n+1}\frac1{3^{i-1}}=\lim_{n\to\infty}\left(\frac32-\frac1{2\cdot3^n}\right)=\frac32

So we have

\displaystyle\sum_{n=1}^\infty30\left(\frac13\right)^{n-1}=30\cdot\frac32=45

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