<h3>Let the number be x</h3>
- Multiply all terms by the same value to eliminate fraction denominators
- Cancel multiplied terms that are in the denominator
<h2>Solved ✔︎</h2>
If you would like to solve the equation x + 1/6 = 6, you can do this using the following steps:
x + 1/6 = 6 /-1/6
x + 1/6 - 1/6 = 6 - 1/6
x = 6 - 1/6
x = 36/6 - 1/6
x = 35/6
x = 5 5/6
The correct result would be C. x = 5 5/6.
A + R + J = 1925
A = 1/2R
J = 2R
1/2R + R + 2R = 1925
7/2R = 1925
R = 550
J = 2(550)
J = 1100
Joel collects 1100 cans.
We evaluate b^2 - 4ac. If the answer is positive, then there are two real solutions. If it is zero, there is one real solution. If it is negative, there are 2 complex solutions. In this equation, a = 1, b = 5, c = 7. Now we plug in the numbers.
Because this answer is positive, there are two real solutions.
Answer:
Part A: Missing Numbers are 64 and 7 going downwards
Part B: Make a TABLE representing how much money they made for 5, 6, 8 hours which they made $64
Part C: ??
Step-by-step explanation:
