Answer:
the answer is 9
Step-by-step explanation:
So distribute using distributive property
a(b+c)=ab+ac so
split it up
(5x^2+4x-4)(4x^3-2x+6)=(5x^2)(4x^3-2x+6)+(4x)(4x^3-2x+6)+(-4)(4x^3-2x+6)=[(5x^2)(4x^3)+(5x^2)(-2x)+(5x^2)(6)]+[(4x)(4x^3)+(4x)(-2x)+(4x)(6)]+[(-4)(4x^3)+(-4)(-2x)+(-4)(6)]=(20x^5)+(-10x^3)+(30x^2)+(16x^4)+(-8x^2)+(24x)+(-16x^3)+(8x)+(-24)
group like terms
[20x^5]+[16x^4]+[-10x^3-16x^3]+[30x^2-8x^2]+[24x+8x]+[-24]=20x^5+16x^4-26x^3+22x^2+32x-24
the asnwer is 20x^5+16x^4-26x^3+22x^2+32x-24
I hope everythink is clearly :) If not just ask :)
answer for your question is starting when x^2-14x-95=0. Just Ignore this above this.
Answer:
(3.5, 17)
Step-by-step explanation:
It would be nice to see the whole graph, so we can see where the functions cross.
Without that information, we can still eliminate unreasonable choices.
A) the quadratic at y=3.5 is well above the exponential
B) the most likely choice (3.5, 17)
C) at x=-8, the quadratic is above the exponential
D) neither graph goes anywhere near y = -8
