1. ![\sqrt[0.4]{29} = 4,528.9236.](https://tex.z-dn.net/?f=%5Csqrt%5B0.4%5D%7B29%7D%20%20%3D%204%2C528.9236.)
2. Jenna bought 4 lbs of apples.
Step-by-step explanation:
Step 1:
If a value x is to be rooted to a value of a, it will be equal to the value of x with an exponential 
.
So ![\sqrt[a]{x} = x^{\frac{1}{a} }](https://tex.z-dn.net/?f=%5Csqrt%5Ba%5D%7Bx%7D%20%3D%20x%5E%7B%5Cfrac%7B1%7D%7Ba%7D%20%7D)
.
In the question, 
.
Substituting these values in the equation we get,
![\sqrt[0.4]{29} = 29^{\frac{1}{0.4} } = 29^{2.5} = 4,528.9236.](https://tex.z-dn.net/?f=%5Csqrt%5B0.4%5D%7B29%7D%20%3D%2029%5E%7B%5Cfrac%7B1%7D%7B0.4%7D%20%7D%20%3D%2029%5E%7B2.5%7D%20%3D%204%2C528.9236.)
Step 2:
Apples cost $2.75 per pound. Jenna paid $11 for apples.
To calculate how many pounds of apples Jenna bought, we divide the price for which apples were bought by the cost per pound.
The price for which apples were bought = $11.
The cost per pound = $2.75.
The number of pounds bought = 
So Jenna bought 4 pounds of apples.
Answer:
No
Step-by-step explanation:
Answer:
= A(t) = 120000(1.06)^t t = 1 year
We just add ^12 to equal 1/12 months.
Then use the notation below.
Step-by-step explanation:
We simply want to write an equivalent form of the same equation that will allow for the time period to be calculated in years.
For one year, t = 1, we want it to grow 6% = 1.06
Yearly Rate of Growth annual equation: = 7200
P(i) 7200/ 120000 x 100% = 6% per year growth rate
the yearly growth factor is 1 + appreciation rate = 1+i =1+0.06= 1.06
So time in years can be added to t
= A(t) = 120000(1.06)^t t = 1 year
The yearly growth factor = 1.06
To equate monthly we add the exponent t*12*1/12 before 120000(1.06)^t
then add 1/12 to replace ^t = monthly, or just keep the t= time
A(t) = 120000(1.06)^12t
Answer:
13
Step-by-step explanation:
56-4=52 then 52/4 is 13
