SVS2652

So do you know who made the website?

How did you figure that out guest!

I bet they are looking for you to use the Pythagorean Theorem. Since the random number will be less than 1, you can assume 1 is the largest side length. For obtuse triangles, a2+b2

now pi/4*4=4pi/4=pi.

So your answer is pi.

Solve the following system:
{3 + 10 d + 5 n = 168 | (equation 1)
5 + 10 d + 5 (n - 1) + p = 168 | (equation 2)
d + n + p = 24 | (equation 3)

Express the system in standard form:
{5 n + 10 d+0 p = 165 | (equation 1)
5 n + 10 d + p = 168 | (equation 2)
n + d + p = 24 | (equation 3)

Subtract equation 1 from equation 2:
{5 n + 10 d+0 p = 165 | (equation 1)
0 n+0 d+p = 3 | (equation 2)
n + d + p = 24 | (equation 3)

Divide equation 1 by 5:
{n + 2 d+0 p = 33 | (equation 1)
0 n+0 d+p = 3 | (equation 2)
n + d + p = 24 | (equation 3)

Subtract equation 1 from equation 3:
{n + 2 d+0 p = 33 | (equation 1)
0 n+0 d+p = 3 | (equation 2)
0 n - d + p = -9 | (equation 3)

Swap equation 2 with equation 3:
{n + 2 d+0 p = 33 | (equation 1)
0 n - d + p = -9 | (equation 2)
0 n+0 d+p = 3 | (equation 3)

Subtract equation 3 from equation 2:
{n + 2 d+0 p = 33 | (equation 1)
0 n - d+0 p = -12 | (equation 2)
0 n+0 d+p = 3 | (equation 3)

Multiply equation 2 by -1:
{n + 2 d+0 p = 33 | (equation 1)
0 n+d+0 p = 12 | (equation 2)
0 n+0 d+p = 3 | (equation 3)

Subtract 2 × (equation 2) from equation 1:
{n+0 d+0 p = 9 | (equation 1)
0 n+d+0 p = 12 | (equation 2)
0 n+0 d+p = 3 | (equation 3)

n = 9 Nickels
d = 12 Dimes
p = 3 Pennies

I feel Post views are decreasing, hopefully, it doesn't end the site, and make it unactive.

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Correct Good Job! How did you figure it out?