$17. (a^{m})^{n}=a^{m*n}\\\\a^{m}*a^{n}=a^{m+n}\\\\-3k^{2}*(4k^{5})^{3}=-3k^{2}*4^{3}*k^{5*3}\\\\ = -3k^{2}*64*k^{15}\\\\= (-3)*64*k^{2+15}\\\\= - 192k^{17}$

$19. \dfrac{a^{m}}{a^{n}}=a^{m-n}\\\\\\\dfrac{(-7p^{4})*(-8p^{5})}{2p^{3}}= \dfrac{(-7)*(-8)*p^{4}*p^{5}}{2p^{3}}\\\\\\=(-7)*(-4)*p^{4+5-3} \\\\= 28p^{6}\\$

$20)\dfrac{15a^{16}b^{11}}{(3a^{4}b^{2})^{3}}=\dfrac{15a^{16}b^{11}}{3^{3}*a^{4*3}*b^{2*3}}\\\\\\=\dfrac{15a^{16}b^{11}}{27*a^{12}*b^{6}}\\\\=\dfrac{5*a^{16-12}*b^{11-6}}{9}\\\\\\=\dfrac{5a^{4}b^{5}}{9}$

4 0

2 years ago

Two cars traveled equal distances in different amounts of time. Car A traveled the distance in 2.4 h, and Car B traveled the dis

Arte-miy333 [17]

One way to go about this is to first list everything we know in the form of variables. This will make it easier to see how these numbers correlate instead of trying to remember formulas to plug these numbers into.

TimeA = 2.4h (time of Car A to travel)
TimeB = 4h (time of Car B to travel)
SpeedA = SpeedB + 22mph (Speed of Car A)
SpeedB = SpeedA - 22mph (Speed of Car B)
Distance = x (the distance traveled by each car)

We are looking for SpeedA. How can we find this? Well, we know that speed multiplied by time is equal to distance, so let's start there.

SpeedA * 2.4h = x
(SpeedB + 22mph) * 2.4h = x
(2.4h * SpeedB) + 52.8miles = x

We also know that:
SpeedB * 4h = x

Since both of these equations are equal to x, we can combine them:
SpeedB * 4h = x = (2.4h * SpeedB) + 52.8miles
SpeedB * 4h = (2.4h * SpeedB) + 52.8miles
1.6h * Speed B = 52.8miles
SpeedB = 52.8/1.6 mph = 33 mph

SpeedA = SpeedB + 22mph = 33mph + 22mph = 55mph

Therefore, Car A was traveling at 55mph.

3 0

3 years ago