Let z be any number (like x would be)
if we set z equal to 2x-6y, then we get z = 2x-6y
The system
2x-6y = 8
2x-6 = 3
turns into this new system
z = 8
z = 3
but z is a single number. It can't be both 8 and 3 at the same time. So there are no solutions
Answer:
B
Step-by-step explanation:
Class A mean: 4.8
Class B mean: 4
B) yes, .8 less
Explanation:
Addition of fractions can be accomplished using the formula ...
a/b + c/d = (ad +bc)/(bd)
Usually, you are asked to find the common denominator and rewrite the fractions using that denominator. It is not necessary, but it can save a step in the reduction of the final result. Here, we'll use the formula, then reduce the result to lowest terms.
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13. 5/6 +9/11 = (5·11 +6·9)/(6·11) = 109/66 = 1 43/66
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14. 7/20 -5/8 = (7·8 -20·5)/(20·8) = -44/160 = -11/40
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15. 1/5 -1/12 = (1·12 -5·1)/(5·12) = 7/60
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Dividing fractions can be accomplished different ways. I was taught to multiply by the inverse of the divisor. ("Invert and multiply.") Here, that means the problem (2/7) / (1/13) can be rewritten as ...
(2/7) × (13/1) . . . . . where 13/1 is the inverse of 1/13.
You can also express the fractions over a common denominator. In that case, the quotient is the ratio of the numerators. Perhaps a little less obvious is that you can express the fractions using a common numerator. Then the quotient is the inverse of the ratio of the denominators: (2/7) / (2/26) = 26/7. (You can see how this works if you "invert and multiply" the fractions with common numerators. Those numerators cancel.)
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16. (2/7)/(1/13) = 2/7·13/1 = 26/7 = 3 5/7
55 cubic units because
- each cube is equal to 1 cubic unit
- the total volume is 60 cubic units
- 5 cubes are already inside
