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defon
3 years ago
14

Find the intergration as 0 approaches 3

Mathematics
1 answer:
Nuetrik [128]3 years ago
6 0
You have to integrate two equations, one for the case where the value in the absolute signs is negative and one for the case where the value is positive...

So you will integrate one on the interval 0 to 1 and the other on the interval from 1 to 3...

⌠-x+1 dx, x=[0,1] + ⌠x-1 dx, x=[1,3]

(-x^2/2+x)[0,1] + (x^2/2-x)[1,3]

(-0.5+1)+(4.5-3-0.5+1)

(0.5)+(2)

2.5 u^2
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If you’re using the inspection method, what constant term would you need in the numerator of this rational expression for (x+5)
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<span>If I am using the inspection method, the constant term that I would need in the numerator of this rational expression for (x+5) to be a common factor of (x^2+6x+14) / (x+5) is 11/(x+1)</span>
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4 years ago
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A cylidrical container of height 28cm and diameter of 18cm is with water. The water is then poured into another container with a
MAVERICK [17]

Answer:

23.9894949495cm is the depth of the container.

Step-by-step explanation:

Cylindrical container:

height: 28

radius: 9 (or 18/2)

pi * 9² = 254.4690049408

254.4690049408 * 28 = 7,125.1321383424 cm²

cylidrical container area: 7,125.1321383424cm²

Container:

length: 27cm

width: 11cm

27 * 11 = 297

7,125.1321383424 / 297 = 23.9894949495cm

I assume that the container with rectangular base is precisely full when tou pour the water into it.

And that your question was how deep the container was.

4 0
4 years ago
Write an equation of a line that is perpendicular to the given line and that passes through the given point.
devlian [24]
Here is the answer I believe. It goes in a x all it did was added to to each x and y (-3,0)
8 0
4 years ago
What are some good electronics that I can buy on Amazon?
Ksivusya [100]

Answer:

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Step-by-step explanation:

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3 0
4 years ago
An open box is to be made out of a 12-inch by 20-inch piece of cardboard by cutting out squares of equal size from the four corn
Natali5045456 [20]

Answer:

Length = 15.14, width = 7.14 and height = 2.43  inches.

( correct to the nearest hundredth).

Step-by-step explanation:

Let the lengths of the squares cut out be x inches.

Then the width of the box will be 12 - 2x and the length will be

20 - 2x inches.

The height of the box is x inches.

Volume V = x(20-2x)(12-2x)

To find the value of x when V is a maximum we find the derivative and equate it to zero.

V = x( 240 - 64x + 4x^2)

V = 4x^3 - 64x^2 + 240x

dV/dx = 12x^2 - 128x + 240 = 0

4(3x^2 - 32x + 60) = 0

This won't factor so we use the formula:

x = [-(-32) +/- √((-32)^2 - 4*3*60)] / 6

= 8.24, 2.43.

So one of these gives a maximum Volume.

The second derivative

d^2V/dx^2 = 24x - 128

When x = 8.24,  this = 69.6 (positive) so this gives a minimum.

x = 2.43 gives a negative  value ( -49.7)  so this is the maximum

So the dimensions are:-

length = 20 - 2(2.43) = 15.14

width = 12 - 2(2.43) = 7.14

height = 2.43.

7 0
3 years ago
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