Isosceles and a right triangle
A. x∧2+(x+20∧2=150∧2 is correct.
because you need to plug the numbers into the pathagoreum theorum which is
a∧2+b∧2=c∧2
150 is our c because it is the side right accross from the right angle.
which number is a and which number is b does not matter.
x is the distance north so we can assign that to a and the distace east is north plus 20 so we can assign b to x+20.
Answer:
1) Combine like terms
2) ![\sqrt[3]{x} =3](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%7D%20%3D3)
3) cube both sides of the equation
4) ![4\sqrt[3]{27} +8\sqrt[3]{27}=36](https://tex.z-dn.net/?f=4%5Csqrt%5B3%5D%7B27%7D%20%2B8%5Csqrt%5B3%5D%7B27%7D%3D36)
5) 4(3) + 8(3) = 36
Step-by-step explanation:
1) Combine like terms
2)
3) cube both sides of the equation
4)
5) 4(3) + 8(3) = 36
Answer:
I think it is A that will take longer but im not 100 percent sure
Step-by-step explanation:
First write an equation:
Location A- 800+100x
Location B- 1200+200x
Then you need to plug numbers in for x
For this case we have the following function:
f (x) = (1/3) * (4 ^ x)
We must evaluate the function for x = 2
We have then:
f (2) = (1/3) * (4 ^ 2)
Rewriting:
f (2) = (1/3) * (16)
f (2) = 16/3
Answer:
The function evaluated at x = 2 is:
f (2) = 16/3
option A
