So, we know that a^2 + b^2 = c^2. Right? That is called the Pythagorean Theorem.
In this case. We can say that 39 is a, 40 is b, and x is c.
NOTE: It doesn't really matter whether 39 is a or b. a & b are just the two legs of the right triangle.
So, if we say that 39 is a, 40 is b, and x is c. We can plug it into the Pythagorean Theorem.
39^2 + 40^2 = x^2
I'll let you take it from there.
0.03125 beciase there are 32 people and one ticket make that as a fraction which is 1/32 and make it a decimal
Answer:
Step-by-step explanation:
We'll take this step by step. The equation is
![8-3\sqrt[5]{x^3}=-7](https://tex.z-dn.net/?f=8-3%5Csqrt%5B5%5D%7Bx%5E3%7D%3D-7)
Looks like a hard mess to solve but it's actually quite simple, just do one thing at a time. First thing is to subtract 8 from both sides:
![-3\sqrt[5]{x^3}=-15](https://tex.z-dn.net/?f=-3%5Csqrt%5B5%5D%7Bx%5E3%7D%3D-15)
The goal is to isolate the term with the x in it, so that means that the -3 has to go. Divide it away on both sides:
![\sqrt[5]{x^3}=5](https://tex.z-dn.net/?f=%5Csqrt%5B5%5D%7Bx%5E3%7D%3D5)
Let's rewrite that radical into exponential form:

If we are going to solve for x, we need to multiply both sides by the reciprocal of the power:

On the left, multiplying the rational exponent by its reciprocal gets rid of the power completely. On the right, let's rewrite that back in radical form to solve it easier:
![x=\sqrt[3]{5^5}](https://tex.z-dn.net/?f=x%3D%5Csqrt%5B3%5D%7B5%5E5%7D)
Let's group that radicad into groups of 3's now to make the simplifying easier:
because the cubed root of 5 cubed is just 5, so we can pull it out, leaving us with:
which is the same as:
![x=5\sqrt[3]{25}](https://tex.z-dn.net/?f=x%3D5%5Csqrt%5B3%5D%7B25%7D)
Answer:
6x = -60
Step-by-step explanation:
-3x - 7 = 23
-3x = 30
x = -10
6(-10) = -60
3x - 12
3(3x/3 - 12/3)
3(x - 4)
The answer is: 3(x - 4).