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 $\boxed{y=8\left(\frac{3}{2}\right)^x}.$

8 0

3 years ago

The product of the slopes of perpendicular lines is always

erastovalidia [21]

The product of the slopes is -1.

8 0

3 years ago