Answer:
15
Step-by-step explanation:
Let n, d, q represent the numbers of nickels, dimes, and quarters. The problem statement tells us ...
n +d +q = 37
n = d +4
q = n +2
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Rearranging the second equation gives ...
d = n -4
Substituting that into the first, we get ...
n + (n -4) +q = 37
2n +q = 41 . . . . . . . add 4 and simplify
Rearranging the third original equation gives ...
n = q -2
Substituting into the equation we just made, we get ...
2(q -2) +q = 41
3q = 45 . . . . . . . . add 4 and simplify
q = 15 . . . . . . . . . divide by 3
Joe has 15 quarters.
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<em>Check</em>
The number of nickels is 2 fewer, so is 13. The number of dimes is 4 fewer than that, so is 9. The total number of coins is 15 + 13 + 9 = 37, as required.
Let the second score = x
⇒ second score = x
<span>the first is 14 points more than the second
</span>⇒ first score = x + 14
<span>the sum of the first two is 6 more than twice the third
</span>⇒ third score = 1/2 (x + x + 14 - 6) = x + 4
<span>The sum of a student's three score is 246
</span>⇒ x + (x + 14) + (x + 4) = 246
<span>
Solve x:
</span>x + x + 14 + x + 4 = 246
3x + 18 = 246
3x = 228
x = 76
second score = x = 76
first score = x + 14 = 76 + 14 = 90
Answer: 90
You could use 10, 6 and 7.
Using ten would have them all equal 50, using 6 would have them all equal 30 and using 7 would have them all equal 35.
Hope this helps!
By just looking at ur whole numbers...12 + 3 + 5 = 20...and thats not even counting ur decimals..so by adding ur decimals, u know that it will be over 20.
<span>Arc length = C*(21/360)
so the answer is D</span>
