A + B + C = 476
A = 3B + 6
C = B + 45
now its just a matter of subbing..
A + B + C = 476
(3B + 6) + B + (B + 45) = 476...combine like terms
5B + 51 = 476
5B = 476 - 51
5B = 425
B = 425/5
B = 85 <== team B scored 85
A = 3B + 6
A = 3(85) + 6
A = 255 + 6
A = 261 <=== team A scored 261
C = B + 45
C = 85 + 45
C = 130 <=== team C scored 130
Answer:
0
Step-by-step explanation:
3/4 - 1/4 = 2/4
2/4 - 1/4
get the same denominators
2/4 /2= 1/8
1/8 - 1/8= 0
There is nothing left in the cereal box.
<h3>x less than or equal to 5</h3>
The average has to be at least 120 and at most 130
To calculate the average we need the sum of all values divided by the number of values, in this case, three (135, 145 and the third result).
120 ≤ (135 + 145 + n)/3 ≤ 130
In inequalities like this, what we change in one side, must be changed in the othe rside as well.
360 ≤ 280 + n ≤ 390
80 ≤ n ≤ 110
Answer:
(b) 1:9
(c) 1:8
Step-by-step explanation:
(b) x*y : kx*ky, so 1:k² with k=3 is 1:9
(c) Assuming the rectangles get a z dimension, the volumes would have a ratio of xyz : xkykzk = xyz : xyzk³ = 1 : k³. With k=2 that is 1:8. But the z was never introduced so this is a bit inconclusive.
