(6x^3/3x^4)^2 = (2/x)^2 = 4/x^2
Set s(t) = 0 which means it hits the ground. The formula doesn't fit the parameters given as it shows that the pitcher is standing on something 37 feet high.
<span>s(t)=-16t^2+140t+37 and has an initial velocity of 140.
Graphing or solving this, t= 9 seconds when it hits the ground.
The velocity V(t) is the derivative of s(t)
V(t) = -32t+140
V(9) = -148 ft/second which is going down
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The answer for this equation is A
Answer:
The integrals was calculated.
Step-by-step explanation:
We calculate integrals, and we get:
1) ∫ x^4 ln(x) dx=\frac{x^5 · ln(x)}{5} - \frac{x^5}{25}
2) ∫ arcsin(y) dy= y arcsin(y)+\sqrt{1-y²}
3) ∫ e^{-θ} cos(3θ) dθ = \frac{e^{-θ} ( 3sin(3θ)-cos(3θ) )}{10}
4) \int\limits^1_0 {x^3 · \sqrt{4+x^2} } \, dx = \frac{x²(x²+4)^{3/2}}{5} - \frac{8(x²+4)^{3/2}}{15} = \frac{64}{15} - \frac{5^{3/2}}{3}
5) \int\limits^{π/8}_0 {cos^4 (2x) } \, dx =\frac{sin(8x} + 8sin(4x)+24x}{6}=
=\frac{3π+8}{64}
6) ∫ sin^3 (x) dx = \frac{cos^3 (x)}{3} - cos x
7) ∫ sec^4 (x) tan^3 (x) dx = \frac{tan^6(x)}{6} + \frac{tan^4(x)}{4}
8) ∫ tan^5 (x) sec(x) dx = \frac{sec^5 (x)}{5} -\frac{2sec^3 (x)}{3}+ sec x
Answer:
Slope= -1
Step-by-step explanation:
Rise over run
2 to the right 2 down = 2/-2
Simplify to -1
