First you need to set up your decimal as a fraction
45 9
---- divided by 5 ------
100 20
9/20 is your answer
Answer:
B
Step-by-step explanation:
Answer:

Step-by-step explanation:
The expression to transform is:
![(\sqrt[6]{x^5})^7](https://tex.z-dn.net/?f=%28%5Csqrt%5B6%5D%7Bx%5E5%7D%29%5E7)
Let's work first on the inside of the parenthesis.
Recall that the n-root of an expression can be written as a fractional exponent of the expression as follows:
![\sqrt[n]{a} = a^{\frac{1}{n}}](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7Ba%7D%20%3D%20a%5E%7B%5Cfrac%7B1%7D%7Bn%7D%7D)
Therefore ![\sqrt[6]{a} = a^{\frac{1}{6}}](https://tex.z-dn.net/?f=%5Csqrt%5B6%5D%7Ba%7D%20%3D%20a%5E%7B%5Cfrac%7B1%7D%7B6%7D%7D)
Now let's replace
with
which is the algebraic form we are given inside the 6th root:
![\sqrt[6]{x^5} = (x^5)^{\frac{1}{6}}](https://tex.z-dn.net/?f=%5Csqrt%5B6%5D%7Bx%5E5%7D%20%3D%20%28x%5E5%29%5E%7B%5Cfrac%7B1%7D%7B6%7D%7D)
Now use the property that tells us how to proceed when we have "exponent of an exponent":

Therefore we get: 
Finally remember that this expression was raised to the power 7, therefore:
[/tex]
An use again the property for the exponent of a exponent:
Answer:
<h2>
AC = 36.01</h2>
Step-by-step explanation:
Given ΔABC and ΔADB, since both triangles are right angled triangles then the following are true.
From ΔADB, AB² = AD²+BD²
Given AB = 24 and AD = 16
BD² = AB² - AD²
BD² = 24²-16²
BD² = 576-256
BD² = 320
BD = 
BD = 17.9
from ΔABC, AC² = AB²+BC²
SInce AC = AD+DC and BC² = BD² + DC² (from ΔBDC )we will have;
(AD+DC)² = AB²+ (BD² + DC²)
Given AD = 16, AB = 24 and BD = 17.9, on substituting
(16+DC)² = 24²+17.9²+ DC²
256+32DC+DC² = 24²+17.9²+ DC²
256+32DC = 24²+17.9²
32DC = 24²+17.9² - 256
32DC = 640.41
DC = 
DC = 20.01
Remember that AC = AD+DC
AC = 16+20.01
AC = 36.01
Answer:
y = 4 + -2x
Step-by-step explanation:
Simplifying
y = -2x + 4
Reorder the terms:
y = 4 + -2x
Solving
y = 4 + -2x
Solving for variable 'y'.
Move all terms containing y to the left, all other terms to the right.
Simplifying
y = 4 + -2x