M(x) = 4x^3 - 5x^2 - 7x
Let us first find the zeros of the function.
That is when it is equal to zero.
m(x) = 4x^3 - 5x^2 - 7x = 0
x(4x^2 - 5x - 7) = 0. Therefore x = 0 or 4x^2 - 5x - 7 = 0.
Using a quadratic function calculator to solve 4x^2 - 5x - 7
x = 2.09, -0.84
Therefore the zeros are x =-0.84, 0, 2.09 for the function m(x).
The intervals observed are imagining that the zeros are on the number line:
x<-0.84, -0.84<x<0, 0<x<2.09, x>2.09.
For each of this range we would test the function with a number that falls in the range.
The function is decreasing in the interval where it is less than 0.
For x<-0.84, let us test x = -1, m(x) = 4x^3 - 5x^2 - 7x = 4(-1)^3 - 5(-1)^2 - 7(-1) = -4 -5 +7 = -2, -2 < 0, so it is decreasing here.
For -0.84<x<0, let us test x = -0.5, m(x) = 4x^3 - 5x^2 - 7x = 4(-0.5)^3 - 5(-0.5)^2 - 7(-0.5) = -0.5 -1.25 +3.5 = 1.75, 1.75 >0. It is not decreasing.
For 0<x<2.09, let us test x = 1, m(x) = 4x^3 - 5x^2 - 7x =
4(1)^3 - 5(1)^2 - 7(1) = 4 -5 -7 = -8, -8 <0. It is decreasing.
For x>2.09, let us test x = 3, m(x) = 4x^3 - 5x^2 - 7x =
4(3)^3 - 5(3)^2 - 7(3) = 108 -45 -21 = 42, 42 >0. It is not decreasing.
So the function is decreasing in the intervals:
x < -0.84, & 0<x<2.09.
Rectangle A ~ Rectangle B
The width of the rectangle is twice the length of the rectangle
<h3>How to compare the dimensions of the rectangle?</h3>
The given parameters are:
Width = 4 cm
Length = 2 cm
Express 4 cm as 2 * 2 cm
Width = 2 * 2 cm
Substitute Length = 2 cm in Width = 2 * 2 cm
Width = 2 * Length
This means that the width of the rectangle is twice the length of the rectangle
Hence, the true statement is that the width of the rectangle is twice the length of the rectangle
Read more about rectangles at:
brainly.com/question/25292087
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Answer:
0 = $0.00
1 = $3.00
2 = $6.00
3 = $9.00
Step-by-step explanation: