Answer:
1. C points radially outward from origin
2. A. Points radially outward from z origin
3. D. Radially inwards toward origin
4. B. Inwards towards the z axis
Step-by-step explanation:
1.
X²+y²+z² = f(x,y,z)
Vector P<x,y,z>
2<x,y,z> = 2p
The answer is c points radially outward from origin
2.
X²+y² = f(x,y)
<2x,2y,0>
Points radially outward from z axis
3.
D. Points radially inwards toward origin
4.
B. Points radially inwards toward the z axis.
Please check attachment for solutions to answers 3 and 4
There's another way to do this, but I'm showing you what I use.
First: I would go ahead and subtract 9x from both sides of the equation. By doing that, the "9x" on the left side would cancel out and you would be left with "7=-6x-5"
Second: Add 5 to both sides of the equation. It should cancel out the 5 on the right and you should be left with "12=-6x"
Lastly: Divide both sides by -6, the -6 on the right next to the variable will cancel out and you'll be left with "-2=x" Which is the solution.
(y+5)(y+5)
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(y-8)(y+5)
y+5 cancels out so the answer is y+5
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y-8
No because all quadrilaterals must equal a total of 360 degrees. If you have four angles that are greater than 90 degrees the sum of all the interior angles is greater than 360 degrees therefore it cannot exist as a quadrilateral.
Answer:
7000
Step-by-step explanation:
