Answer:
all work is shown and pictured
Answer:
-t2+10t-h+8 = 0
Step-by-step explanation:
<span>2x - 3y - 3z = 8
We've been given the normal vector to the plane <2, -3, -3> and a point within the plane (1, -5, 3). In general if you've been given both the normal vector <a,b,c> and a point (e,f,g) within the plane, the expression for the plane will be:
ax + by + cz = d
and you can compute d by:
d = ae + bf + cg
So let's calculate d:
d = ae + bf + cg
d = 2*1 + -3*-5 + -3*3
d = 2 + 15 + -9
d = 8
And the equation for the plane is
2x - 3y - 3z = 8</span>
Answer:
The area of sector AOB is 75.36 unit²
Step-by-step explanation:
Consider the provided information.
Circle O with minor arc AB=60 degrees and OA =12.
We need to find the area of sector.
Area of the sector: 
Substitute 
and r =12 in above formula.
Hence, the area of sector AOB is 75.36 unit²
Positive 2 is your slope, (rise over run) if you have to graph it and -1 is your y-intercept
