Answer:
D. ![\sqrt[3]{5^7}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B5%5E7%7D)
Step-by-step explanation:
The formula for this conversion is:
![a^\frac{x}{n}=\sqrt[n]{a^x}](https://tex.z-dn.net/?f=a%5E%5Cfrac%7Bx%7D%7Bn%7D%3D%5Csqrt%5Bn%5D%7Ba%5Ex%7D)
Substitute the values you currently have.
![5^\frac{7}{3}=\sqrt[n]{a^x}](https://tex.z-dn.net/?f=5%5E%5Cfrac%7B7%7D%7B3%7D%3D%5Csqrt%5Bn%5D%7Ba%5Ex%7D)
Since we know that
a = 5
x = 7
n = 3
Fill the square root with this.
![5^\frac{7}{3}=\sqrt[3]{5^7}](https://tex.z-dn.net/?f=5%5E%5Cfrac%7B7%7D%7B3%7D%3D%5Csqrt%5B3%5D%7B5%5E7%7D)
The answer is n+7 and 7+n
<h2>
Explanation:</h2><h2>
</h2>
Here we know that Margie is standing on a bridge looking at the river. She throws a rock over the bridge with an initial velocity of 60 ft/sec. The height, h, of the rock at the rime t seconds after it was thrown can be modeled by the equation:

To calculate how long will it take for the rock to reach the surface of the water, we need to set
. So:

Given that time can't be negative, we just choose the positive value. Therefore, <em>it will take 4.70 seconds for the rock to reach the surface water.</em>
Answer:
$14.50 I'm positive.
Step-by-step explanation:
Rented one time shoes: $2.75
One game of bowling: $2.50*4 games of bowling
4 games of bowling is $10
One nachos: $1.75*2 orders of nachos
2 orders of nachos is $3.50
- Half price of nachos, so 3.50/2 which is 1.75.
- $2.75+$10+$1.75= <u>$14.50</u>
HOPE THIS HELPED :)
Answer:
Wanda and Dave will catch each other in 54 seconds after Dave starts walking.
Step-by-step explanation:
Let Wanda and Dave catch each other when x be the time after Dave starts walking and y be the distance covered by them
It is given that Wanda started walking along a path 27 seconds before Dave and the constant speed of Wanda is 3 feet per second.



.... (1)
The constant speed of Dave is 4.5 feet per second.

.... (2)
Equate equation (1) and (2).


Divide both sides by 1.5.


Therefore, Wanda and Dave will catch each other in 54 seconds after Dave starts walking.