To solve for the time an object takes to hit the ground, the equation
<em>y</em> <em>= y₀ + v₀t + ½gt²</em>, or just <em>0 = y₀ +½gt²</em> since there is no initial velocity and the final y position is 0 (since it is on the ground). Therefore, <em>t = (−2y₀/g)^½.</em>
(-2*60/-9.8)^½ = 3.50 seconds.
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Let s represent the short side of the triangle. The long sides of the triangle are each s+1, and the triangle's perimeter is
... s + (s+1) + (s+1) = 3s+2
The length of one side of the square is s-2, and its perimeter is 4 times that, 4(s-2) = 4s-8. The square and triangle have the same perimeter, so
... 3s+2 = 4s-8
... 10 = s . . . . . . . . add 8-3s to both sides
The length of the shorter side of the triange is 10 units.
That is the slope at f(1)
one way is to just take the derivitive
f'(x)=2x+(1/x)
for ti-89, do calc then derivitive then input "x^2+ln(x),x)" and click enter to get the derivitive
you would get f'(x)=2x+(1/x)
an easier way would be to have
click calc then differntiate then type x^2+ln(x),x)|x=1 then click enter
you get f'(1)=3
answer is 3
12 meters because 100 centimeters is equal to one meter.
