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

4 0

3 years ago

Write an exponential equation in the form y = ab* whose graph passes through points (2, 18) and (5, 60.75).

DochEvi [55]

Let $y=ab^x,$ as described in the problem statement. Plugging in $x=2$ and $x=5$ gives us the equations $18=ab^2$ and $60.75=ab^5.$ Since $a,b$ are nonzero, we divide the second equation by the first to get $\frac{27}{8}=b^3,$ which means that $b=\frac{3}{2},$ after taking the cube root. Plugging this value for $b$ into the first equation, we now have $18=a(3/2)^2=\frac{9a}{4}.$ Solving gives $a=8,$ and our desired exponential equation is $\boxed{y=8\left(\frac{3}{2}\right)^x}.$