Answer:
Step-by-step explanation:
did u try seaching the answer?
9514 1404 393
Answer:
PJ ⊥ KP
Step-by-step explanation:
The little square corner at point P is the mark that indicates lines are perpendicular.
That is the only on mark indicating perpendicular in this diagram. It shows line KI to be perpendicular to line MJ. Then all of the segments on one of those lines are perpendicular to any of the segments on the other of those two lines.
Just to make sure we don't get in trouble with interpretation, we should probably include point P in any segments we list.
MJ ⊥ KI
OJ ⊥ KI
MP ⊥ KI
OP ⊥ KI
PJ ⊥ KI
In any of the above, KI can be replaced by KP or PI and the statement will remain true.
_____
<em>Additional comment</em>
Segment NI, being parallel to OP, is also perpendicular to KI. However, it is not marked as such.
MO is perpendicular to PI, but isn't really marked as such.
No 7/8 is lager then 2/3.
2/3 = 0.6666
7/8= 0.87..
You need three groups of bananas, one group of raspberries, and one group of strawberries.
Hope this helps.
Answer:
![f(x_1)=2\sqrt{118}\\f(x_2)=-2\sqrt{118}](https://tex.z-dn.net/?f=f%28x_1%29%3D2%5Csqrt%7B118%7D%5C%5Cf%28x_2%29%3D-2%5Csqrt%7B118%7D)
Step-by-step explanation:
![f(x)=13x^3-x^2-3x+10](https://tex.z-dn.net/?f=f%28x%29%3D13x%5E3-x%5E2-3x%2B10)
Differentiate with respect to ![x](https://tex.z-dn.net/?f=x)
![f'(x)=39x^2-2x-3](https://tex.z-dn.net/?f=f%27%28x%29%3D39x%5E2-2x-3)
For equation of the form
, solutions are given by ![x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B-b%5Cpm%20%5Csqrt%7Bb%5E2-4ac%7D%7D%7B2a%7D)
To find: critical points
![39x^2-2x-3=0](https://tex.z-dn.net/?f=39x%5E2-2x-3%3D0)
![x=\frac{2\pm \sqrt{4+468}}{78}\\=\frac{2\pm \sqrt{472}}{78}\\=\frac{2\pm 2\sqrt{118}}{78}\\=\frac{1\pm \sqrt{118}}{39}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B2%5Cpm%20%5Csqrt%7B4%2B468%7D%7D%7B78%7D%5C%5C%3D%5Cfrac%7B2%5Cpm%20%5Csqrt%7B472%7D%7D%7B78%7D%5C%5C%3D%5Cfrac%7B2%5Cpm%202%5Csqrt%7B118%7D%7D%7B78%7D%5C%5C%3D%5Cfrac%7B1%5Cpm%20%5Csqrt%7B118%7D%7D%7B39%7D)
Let ![x_1=\frac{1+\sqrt{118}}{39}\,,\,x_2=\frac{1- \sqrt{118}}{39}](https://tex.z-dn.net/?f=x_1%3D%5Cfrac%7B1%2B%5Csqrt%7B118%7D%7D%7B39%7D%5C%2C%2C%5C%2Cx_2%3D%5Cfrac%7B1-%20%5Csqrt%7B118%7D%7D%7B39%7D)
Differentiate
again with respect to ![x](https://tex.z-dn.net/?f=x)
![f''(x)=78x-2](https://tex.z-dn.net/?f=f%27%27%28x%29%3D78x-2)
![f(x_1)=f\left (\frac{1+\sqrt{118}}{39} \right )\\=78\left ( \frac{1+\sqrt{118}}{39} \right )-2\\=2\left ( 1+\sqrt{118} \right )-2\\=2\sqrt{118}\\f(x_2)=f\left (\frac{1-\sqrt{118}}{39} \right )\\=78\left ( \frac{1-\sqrt{118}}{39} \right )-2\\=2\left ( 1-\sqrt{118} \right )-2\\=-2\sqrt{118}](https://tex.z-dn.net/?f=f%28x_1%29%3Df%5Cleft%20%28%5Cfrac%7B1%2B%5Csqrt%7B118%7D%7D%7B39%7D%20%20%5Cright%20%29%5C%5C%3D78%5Cleft%20%28%20%5Cfrac%7B1%2B%5Csqrt%7B118%7D%7D%7B39%7D%20%5Cright%20%29-2%5C%5C%3D2%5Cleft%20%28%201%2B%5Csqrt%7B118%7D%20%5Cright%20%29-2%5C%5C%3D2%5Csqrt%7B118%7D%5C%5Cf%28x_2%29%3Df%5Cleft%20%28%5Cfrac%7B1-%5Csqrt%7B118%7D%7D%7B39%7D%20%20%5Cright%20%29%5C%5C%3D78%5Cleft%20%28%20%5Cfrac%7B1-%5Csqrt%7B118%7D%7D%7B39%7D%20%5Cright%20%29-2%5C%5C%3D2%5Cleft%20%28%201-%5Csqrt%7B118%7D%20%5Cright%20%29-2%5C%5C%3D-2%5Csqrt%7B118%7D)