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AlladinOne [14]
4 years ago
11

Can someone pls help me answer this, step by step would be helpful but u don’t have to. ty! :)

Mathematics
1 answer:
kipiarov [429]4 years ago
6 0

Answer:

The area is: 4x^5 + 8x^3

Equation: (2x^3 + 4x) * 2x^2

^ = exponent

* = multiply

Step-by-step explanation:

When you try to find the rectangle of an area, the formula is:

Length * Width

In this case, the length is:

2x^3 + 4x

And the width is:

2x^2

Now let's solve the equation.

(2x^3 + 4x) * 2x^2

We need to combine like terms

2x^3 + 2x^2 = 4x^5

After evaluating the equation:

4x^5 + 8x^3

The area is: 4x^5 + 8x^3

hope this helped :)

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azamat

Answer:

101 degrees

Step-by-step explanation:

the sum of all angles in a triangle is always 180 degrees.

the other not-shaded angle in the lower triangle is also 42 degrees.

the shaded angle is therefore

180 - 37 - 42 = 101 degrees

4 0
4 years ago
What is the value of x x=m x-3
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If x=mx-3 (which is not clear from your input),
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then x-mx = -3, or mx-1x = 3, and x(m-1) = 3, and so x = ---------
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4 years ago
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Step-by-step explanation:

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8 0
3 years ago
Determine the standard form of the equation of the line that passes through (6, -9) and (-8, 6)
horrorfan [7]

9514 1404 393

Answer:

  A.  15x +14y = -36

Step-by-step explanation:

Since we are given two points, we can start with the 2-point form of the equation for a line.

  y = (y2 -y1)/(x2 -x1)(x -x1) +y1

  y = (6 -(-9))/(-8 -6)(x -6) +(-9)

  y = 15/-14(x -6) -9

Multiplying by -14, we have ...

  -14y = 15x -90 +126

Adding 14y-36 to both sides gives ...

  -36 = 15x +14y . . . . matches choice A

The standard-form equation is ...

  15x +14y = -36

_____

<em>Additional comments</em>

It can be easier to start with the form ...

  (Δy)x -(Δx)y = (Δy)x1 -(Δx)y1 . . . . . where Δx = x2-x1 and Δy = y2-y1

This gives ...

  (6+9)x -(-8-6)y = 15(6) +14(-9)

  15x +14y = -36 . . . simplified

__

You can also start with the slope-intercept form or the point-slope form, if you're more familiar with those. The result will be the same. I find it handy to be familiar with a number of different forms of the equation for a line.

4 0
3 years ago
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