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3 years ago

5

Simplify the expression.(4+2i)-(1-i)

1 answer:

tester [92]3 years ago

3 0

ANSWER :

6i - (1-i)

Step - by - step explanation:

( 4 + 2i ) - ( 1 - i )

( 4 + 2 × i) - ( 1 - i )

( 6× i ) - ( 1 - i )

= 6i- (1-i)

Hope this helps and pls mark as brainliest :)

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Answer:

The integral of the volume is:

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Step-by-step explanation:

Given

Curve: $x^2 + 4y^2 = 4$

About line $x = 2$ --- Missing information

Required

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$x^2 + 4y^2 = 4$

Make x^2 the subject

$x^2 = 4 - 4y^2$

Square both sides

$x = \sqrt{(4 - 4y^2)$

Factor out 4

$x = \sqrt{4(1 - y^2)$

Split

$x = \sqrt{4} * \sqrt{(1 - y^2)$

$x = \±2 * \sqrt{(1 - y^2)$

$x = \±2 \sqrt{(1 - y^2)$

Split

$x_1 = -2 \sqrt{(1 - y^2)}\ and\ x_2 = 2 \sqrt{(1 - y^2)}$

Rotate about x = 2 implies that:

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So:

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$r_1 = 2 +2 \sqrt{(1 - y^2)}$

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Using washer method along the y-axis i.e. integral from 0 to 1.

We have:

$V = 2\pi\int\limits^1_0 {(r_1^2 - r_2^2)} \, dy$

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$V = 2\pi\int\limits^1_0 {(( 2 +2 \sqrt{(1 - y^2)})^2 - ( 2 -2 \sqrt{(1 - y^2)})^2)} \, dy$

Evaluate the squares

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Remove brackets and collect like terms

$V = 2\pi\int\limits^1_0 {4 - 4 + 8\sqrt{(1 - y^2)} +8 \sqrt{(1 - y^2)}+ 4(1 - y^2) - 4(1 - y^2)} \, dy$

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Rewrite as:

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Using the calculator:

$\int\limits^1_0 {\sqrt{(1 - y^2)} \, dy = \frac{\pi}{4}$

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$V =8\pi^2$

Take:

$\pi = 3.142$

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