Find the length of the hypotenuse. a=20 in b=21in c=? SHOW WORK
2 answers:
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Solve the following system:
{7 x - 3 y = 37 | (equation 1)
{14 x + 11 y = 23 | (equation 2)
Swap equation 1 with equation 2:
{14 x + 11 y = 23 | (equation 1)
{7 x - 3 y = 37 | (equation 2)
Subtract 1/2 × (equation 1) from equation 2:
{14 x + 11 y = 23 | (equation 1)
{0 x - (17 y)/2 = 51/2 | (equation 2)
Multiply equation 2 by 2/17:
{14 x + 11 y = 23 | (equation 1)
{0 x - y = 3 | (equation 2)
Multiply equation 2 by -1:
{14 x + 11 y = 23 | (equation 1)
{0 x+y = -3 | (equation 2)
Subtract 11 × (equation 2) from equation 1:
{14 x+0 y = 56 | (equation 1)
{0 x+y = -3 | (equation 2)
Divide equation 1 by 14:
{x+0 y = 4 | (equation 1)
{0 x+y = -3 | (equation 2)
Collect results:
Answer: {x = 4
{y = -3
{7 x - 3 y = 37 | (equation 1)
{14 x + 11 y = 23 | (equation 2)
Swap equation 1 with equation 2:
{14 x + 11 y = 23 | (equation 1)
{7 x - 3 y = 37 | (equation 2)
Subtract 1/2 × (equation 1) from equation 2:
{14 x + 11 y = 23 | (equation 1)
{0 x - (17 y)/2 = 51/2 | (equation 2)
Multiply equation 2 by 2/17:
{14 x + 11 y = 23 | (equation 1)
{0 x - y = 3 | (equation 2)
Multiply equation 2 by -1:
{14 x + 11 y = 23 | (equation 1)
{0 x+y = -3 | (equation 2)
Subtract 11 × (equation 2) from equation 1:
{14 x+0 y = 56 | (equation 1)
{0 x+y = -3 | (equation 2)
Divide equation 1 by 14:
{x+0 y = 4 | (equation 1)
{0 x+y = -3 | (equation 2)
Collect results:
Answer: {x = 4
{y = -3
1 mi = 5280 ft
1 h = 3600 s
therefore:
Answer: 31.8 mi/h = 46.64 ft/s
1 h = 3600 s
therefore:
Answer: 31.8 mi/h = 46.64 ft/s