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Licemer1 [7]
2 years ago
9

What is the inequality that represents the range of the function g(x)=-3/4x^2 when the domain is restricted to -4

Mathematics
1 answer:
enyata [817]2 years ago
4 0

Answer:

D

Step-by-step explanation:

to show the general direction of the graph, we should find the end points and one in the middle, lets see what happens why x is -4, 0, and 4. when x=-4, it is 16*-3/4 (which is negetive), when x is 0, y is 0, and when x is 4, the same this happens in the first number. If these points are plotted, it is shown as a parabola pointing downwards. The max is will be is 0. because before 0, the numbers are negetive in the y axis, and after they are negetive in the y axis, and it tops out at y=0 when x=0. Therefore Y is smaller than or equal to 0, or D.

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The volume of a rectangular prism is b3 + 8b2 + 19b + 12 cubic units, and its height is b + 3 units. The area of the base of the
omeli [17]
Given:

The volume of the rectangular prism is 

b^{3}+8 b^{2} +19b+12,

the height is h=(b+3)

1. The volume of a rectangular prism is (base area)*height

also, notice that the volume is a third degree polynomial, the height is a 1st degree polynomial, so the base area must be a 2nd degree polynomial, whose coefficients we don't know yet.
Let this quadratic polynomial be (mb^{2}+nb+k)

 
2

b^{3}+8 b^{2} +19b+12=(mb^{2}+nb+k)*(b+3)


notice that  b^{3} is the product of the largest 2 terms: mb^{2} and b, so m must be 1

also, notice that 12 is the product of the constants, k and 3

so k*3=12, this means k=4

3
we write the above equality again:

b^{3}+8 b^{2} +19b+12=(b^{2}+nb+4)*(b+3)


(b^{2}+nb+4)(b+3)= b^{3}+3 b^{2} +nb^{2}+3nb+4b+12

== b^{3}+(n+3)b^{2}+(3n+4)b+12


4
now compare the coefficient with the left side:

8 b^{2}=(n+3)b^{2}

8=n+3

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substituting n=5: 

the base area is b^{2}+5b+4

Answer: b^{2}+5b+4


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3 years ago
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