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Alborosie
3 years ago
12

Click an item in the list or group of pictures at the bottom of the problem and, holding the button down; drag it into the corre

ct position in the answer box. Release your mouse button when the item is place.
If you change your mind, drag the item to the trashcan. Click the trashcan to clear all your answers.
Add the following polynomials, and then place the answer in the proper location on the grid. Write the answer in descending powers of x. Find the sum of (5x3 + 3x2 - 5x + 4) and (8x3 -5x2 + 8x + 9).
Mathematics
1 answer:
Alex777 [14]3 years ago
3 0
8x^3 -5x^2 + 8x + 9+5x^3 + 3x^2 - 5x + 4 = 
8x^3+5x^3-5x^2 + 3x^2+ 8x - 5x+ 9<span>+ 4 = </span>
13x^3-2x^2+3x+13
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Q5. Area of circle=\pi * ( \frac{6}{2} )^{2}
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3 years ago
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Two large and 1 small pumps can fill a swimming pool in 4 hours. One large and 3 small pumps can also fill the same swimming poo
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7 0
2 years ago
The length of the rectangle is twice its width. Write and solve a system of linear equations to find the length L and width W of
vazorg [7]

Answer:

The length is equal to 12 and the width is equal to 6.

Step-by-step explanation:

In order to find the values here, we start by setting the width equal to x. Now knowing this, we know that the length is twice that long. Therefore, the length would be equal to 2x. Now we can use the perimeter formula to solve the equation.

2L + 2W = P

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3 years ago
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In a 45 45 90 right triangle the length of the hypotenuse is 15/2 what is length on one of the legs
Nikitich [7]
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For any 45 45 90 right triangle, the length of the hypotenuse is always the square root of 2 times one of the legs.

To get the lengths of the leg, let's divide the length of the hypotenuse by the square root of 2.

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When simplifying fractions like this, we want to avoid having radicals on the bottom. In order to remove the radical from the denominator, let's multiply the numerator and the denominator by the square root of 2.

We get the following:

\frac{15 \sqrt{2} }{4}

That's your answer.

Have an awesome day! :)

- collinjun0827, Junior Moderator
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3 years ago
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