$3^d\cdot\dfrac{\sqrt5}{3^2\cdot\sqrt{45}}=3\\\\\dfrac{3^d\sqrt5}{3^2\sqrt{9\cdot5}}=3\qquad\text{use}\ \sqrt{ab}=\sqrt{a}\cdot\sqrt{b}\\\\\dfrac{3^d\sqrt5}{3^2\sqrt9\cdot\sqrt5}=3\\\\\dfrac{3^d}{3^2\cdot3}=3\qquad\text{use}\ a^n\cdot a^m=a^{n+m}\\\\\dfrac{3^d}{3^{2+1}}=3\\\\\dfrac{3^d}{3^3}=3\qquad\text{multiply both sides by}\ 3^3\\\\3^d=3\cdot3^3\qquad\text{use}\ a^n\cdot a^m=a^{n+m}\\\\3^d=3^{1+3}\\\\3^d=3^4\iff\boxed{d=4}$

8 0

2 years ago

$\mbox{a rocket is shot straight into the air at an initial velocity of 96 feet/second. }$ $\mbox{the height of the rocket after

Stolb23 [73]

?????? what even is that question

4 0

2 years ago

A frame is 6in tall and 8 in wide. What if it was reduced to a width of 4in then how tall will it be

kkurt [141]

Answer:

3 in

Step-by-step explanation:

Apparently, we are to assume that the height and width are proportional:

height/width = (6 in)/(8 in) = (new height)/(4 in)

Multiplying by 4 in will tell us the new height:

new height = (4 in)(6/8) = 3 in

The reduced frame will be 3 inches tall.

6 0

2 years ago