Answer:
the length of the hypotenuse must be 10.
Step-by-step explanation:
This is a right triangle, so we can apply the Pythagorean Theorem.
6² + 8² = 10² so the length of the hypotenuse must be 10.
Answer:
Equation of tangent plane to given parametric equation is:

Step-by-step explanation:
Given equation
---(1)
Normal vector tangent to plane is:


Normal vector tangent to plane is given by:
![r_{u} \times r_{v} =det\left[\begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\cos(v)&sin(v)&0\\-usin(v)&ucos(v)&1\end{array}\right]](https://tex.z-dn.net/?f=r_%7Bu%7D%20%5Ctimes%20r_%7Bv%7D%20%3Ddet%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%5Chat%7Bi%7D%26%5Chat%7Bj%7D%26%5Chat%7Bk%7D%5C%5Ccos%28v%29%26sin%28v%29%260%5C%5C-usin%28v%29%26ucos%28v%29%261%5Cend%7Barray%7D%5Cright%5D)
Expanding with first row

at u=5, v =π/3
---(2)
at u=5, v =π/3 (1) becomes,



From above eq coordinates of r₀ can be found as:

From (2) coordinates of normal vector can be found as
Equation of tangent line can be found as:

I think b=6. so try that ok
Answer:




Step-by-step explanation:

subtract 24 from both sides

divide both sides by 4

---------->>>>

combine like terms

divide both sides by 6

---------->>>>

add 72 to both sides

divide both sides by 9

---------->>>>

combine like terms

divide both sides by 8

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Hope this is helpful.
Answer:
3y= -2x-6
Step-by-step explanation:
The two points on the line are (-3,0) and (0,-2)
so you first get the gradient;
gradient= <u>change in y</u>
change in x
= <u>-2-0</u>
0-(-3)
=<u> -2</u>
3
so the answer above is the gradient
then pick one point that you used to get the gradient with, so as for me I'll pick (-3,0) and then a general point which is always (x, y)
since you have the gradient you can easily get the equation by doing this
<u>y-0</u><u>. </u><u> </u> = <u>-2</u>
x-(-3). 3
then crossmultiply to get the equation of the line
3y= -2(x+3)
3y= -2x-6