Answer:
(-2,2)
Step-by-step explanation:
x = number of 1-cent stamps
y = number of 8-cent stamps
z = number of 12-cent stamps
We have 31 stamps all together, so x+y+z = 31.
"I have 4 more 1-cent stamps than 8-cent stamps" means we have the equation x = y+8. Whatever y is, add 8 to it to get x. Solve for y to get y = x-8.
You also have "twice as many one cent stamps as 12 cent stamps", so x = 2z. Solving for z gets you z = 0.5x
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x+y+z = 31
x+x-8+z = 31 ... y replaced with x-8
x+x-8+0.5x = 31 ... plug in z = 0.5x
2.5x-8 = 31
2.5x = 31+8
2.5x = 39
x = 39/2.5
x = 15.6
Your teacher made a typo somewhere because we should get a positive whole number result for x (since x is a count of how many 1-cent stamps we have).
The given in the problem above is an arithmetic sequence with the first term equal to 4 and the common difference is 5. To determine the number of seats on row 23, use the formula, an = a1 + (n - 1) d
Solving for the 23rd term, an = 4 + (22) 5 = 114 seats
Therefore, the answer is there are 114 seats on the 23rd row.
Welcome Bby (;
Answer:
ok so the two other ways are- 6:2 and 9:3
And the second part- 5
Step-by-step explanation:
