SOLUTION: 1. the larger leg of a right triangle is 3cm longer than its smaller leg. The hypotenuse is 6 cm longer than the smaller leg. How many centimeters long is the smaller leg?
2. t
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-> SOLUTION: 1. the larger leg of a right triangle is 3cm longer than its smaller leg. The hypotenuse is 6 cm longer than the smaller leg. How many centimeters long is the smaller leg?
2. t
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Question 1031362: 1. the larger leg of a right triangle is 3cm longer than its smaller leg. The hypotenuse is 6 cm longer than the smaller leg. How many centimeters long is the smaller leg?
2. the function f(t)= -5t^2+20t+60 models the approximate height of an object t seconds after its launched. How many seconds does it take the object to hit the ground?
You can put this solution on YOUR website! 1. s = length of smaller leg
l = larger leg
h = hypotenuse
l = s + 3
h = s + 6
substituting
s^2 + s^2 + 6s + 9 = s^2 + 12s + 36
add -s^2 to each side
s^2 + 6s + 9 = 12s + 36
add -12s to each side
s^2 - 6s + 9 = 36
add -36 to each side
s^2 - 6s - 27 = 0
(s - 9)(s + 3) = 0
So s = 9 , s cannot be -3 since this would be a negative length.
2.
f(t)= -5t^2+20t+60
We want to find the values for t where f(t) = 0 .
0 = -5t^2+20t+60
divide each side by -5
0 = t^2 - 4t - 12
0 = (t - 6)(t + 2)
So t = 6. t = -2 does not work since we cannot have negative time.
Confirm by substituting in our original equation.
0 = -5(6)^2+20(6)+60
0 = -5(36)+20(6)+60
0 = -180 +120 + 60
0 = 0
So it checks out.
