<span>There are three possibilities.
1. Of the 30 bikes that needed gear repairs, 20 also needed tires.
2. Of the 30 bikes that needed gear repairs, 10 also needed tires.
3. Of the 30 bikes that needed gear repairs, between 10 and 20 also needed tires.</span>
I'd suggest using "elimination by addition and subtraction" here, altho' there are other approaches (such as matrices, substitution, etc.).
Note that if you add the 3rd equation to the second, the x terms cancel out, and you are left with the system
- y + 3z = -2
y + z = -2
-----------------
4z = -4, so z = -1.
Next, multiply the 3rd equation by 2: You'll get -2x + 2y + 2z = -2.
Add this result to the first equation. The 2x terms will cancel, leaving you with the system
2y + 2z = -2
y + z = 4
This would be a good time to subst. -1 for z. We then get:
-2y - 2 = -2. Then y must be 0. y = 0.
Now subst. -1 for z and 0 for y in any of the original equations.
For example, x - (-1) + 3(0) = -2, so x + 1 = -2, or x = -3.
Then a tentative solution is (-3, -1, 0).
It's very important that you ensure that this satisfies all 3 of the originale quations.
V = a²h/3
Plugging values in we get 8.67X10^7, or 86,700,000
Answer:
No, the student is not correct. The correct expression is 9(n+13).
Step-by-step explanation:
"9 times the sum of a number and 13" means you must multiply 9 by the sum, not just by the number. 9n + 13 would be "the sum of 9 times a number and 13." Slightly different wording.
