Answer:
or
Step-by-step explanation:
yz-plane has the equation 
where 
is a real constant. The normal vector of this plane (vector perpendicular to the plane) is
If the line is perpendicular to this plane, then it is parallel to the normal vector, so the equation of this line is
where 
are coordinates of the point the line is passing through.
In your case, the line is passing through the point (0,0,8), so the canonical equation of the line is
Write a vector parametrization for this line
Answer:
Step-by-step explanation:
Answer:
gabby is correct
Step-by-step explanation:
Answer:
y = 1/2
x = -3/2
Step-by-step explanation:
y = 3x+5
Substance instances of y in the equation.
x + 3x +5 = -1
4x=-6
x= -6/4
y = 1/2
First, let's get the area of the entire triangle since we'll need it later. The area of a triangle is A=1/2*b*h
We can find the height with the Pythagorean theorem by splitting the triangle in half.
3^2+b^2=6^2
9+b^2=36
b^2=27
b=√27
Then we can find the area:
A=1/2*6*√27
A=3√27 or =9√3 or 15.59
Now we can find the area of each region in the triangle other than the shaded region because they are all portions of a circle.
Each region has an angle of 60 because this is an equilateral triangle. Therefore the area of each region other than the shaded region will be 1/6 the area of a circle with a radius of 3 because a full circle is 360 degrees.
A=pi*r^2/6
A=pi*9/6
A=4.71
So three of these regions would have an area of 14.14
We do the area of the triangle minus the area of these regions to get the area of the shaded region
15.59-14.14 = 1.45
Hope this helps!
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