Answer:
idk
Step-by-step explanation: sorry im in unit 6 in science im working on rn
baiiii
Answer:
Step-by-step explanation:
ABC have sides: 5, 7 and 10
5^2 + 7^2 = 25+47 = 72 < 10^2
so triangle ABC is obtuse
JKL has sides: 12, 35 and 37
12^2 + 35^2 = 144 + 1225 = 1369 = 37^2
so triangle JKL is right-angled
PQR has sides 12, 10 and 16
12^2 + 10^2 = 144 + 100 = 244 > 16^2
so triangle PQR is acute
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:
8(6-x)
Step-by-step explanation:
Both 48 and 8 can be divisible by 8.
48 ÷ 8 = 6
8 ÷ 8 = 1
Therefore you get the answer 8(6-x)
as the simplest form.
Hope this helps.
P: quadrilaterals . I hope that helps