Answer:
answer is -6=9
Step-by-step explanation:
Triangle area= b*h/2
Plug in the values:
25*8/2
200/2
100
Final answer: 100 ft^2
Answer:
the answer to your question is (x+2) * (x-12)
The point (2, 5) is not on the curve; probably you meant to say (2, -5)?
Consider an arbitrary point Q on the curve to the right of P,
, where
. The slope of the secant line through P and Q is given by the difference quotient,
![\dfrac{(t^2-2t-5)-(-5)}{t-2}=\dfrac{t^2-2t}{t-2}=\dfrac{t(t-2)}{t-2}=t](https://tex.z-dn.net/?f=%5Cdfrac%7B%28t%5E2-2t-5%29-%28-5%29%7D%7Bt-2%7D%3D%5Cdfrac%7Bt%5E2-2t%7D%7Bt-2%7D%3D%5Cdfrac%7Bt%28t-2%29%7D%7Bt-2%7D%3Dt)
where we are allowed to simplify because
.
Then the equation of the secant line is
![y-(-5)=t(x-2)\implies y=t(x-2)-5](https://tex.z-dn.net/?f=y-%28-5%29%3Dt%28x-2%29%5Cimplies%20y%3Dt%28x-2%29-5)
Taking the limit as
, we have
![\displaystyle\lim_{t\to2}t(x-2)-5=2(x-2)-5=2x-9](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Clim_%7Bt%5Cto2%7Dt%28x-2%29-5%3D2%28x-2%29-5%3D2x-9)
so the slope of the line tangent to the curve at P as slope 2.
- - -
We can verify this with differentiation. Taking the derivative, we get
![\dfrac{\mathrm dy}{\mathrm dx}=2x-2](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cmathrm%20dy%7D%7B%5Cmathrm%20dx%7D%3D2x-2)
and at
, we get a slope of
, as expected.