Cube of x is x^3, according to question
Y= 1/4 x ^3 -2
Answer:
you would have 1,194.4
Step-by-step explanation:
i subtracted.
For #4, we know V = lwh
We have the values
5184 = 2(18)h
Solve for h
5184 = 36h
144 = h.
for # 3:
we know A = pi r^2
We have
A = 3.14 * 6^2
A = 3.14 * 36
A = 113.04
for #2:
We know C = 2pi *r
C = 2(3.14)(14)
C = 28(3.14)
C = 87.92
for #1:
C = 2pi * r
C = 2(3.14)(26.3/2)
C = 2(3.14)(13.15)
C = 26.3(3.14)
C = 82.582
We either need to see a picture of this and/or get more information about the measurements of the triangle. In general, the area outside of the triangle will be the area of the semi-circle minus the area of the triangle itself, or: 1/2*49*3.14 - 1/2 b*h of the triangle. That first part, which is the area of the semi-circle, works out to 76.93 So based on the info we have, it becomes 76.93 - 1/2*b*h of the triangle = area outside of triangle.
Answer:
See Explanation;
Step-by-step explanation:
The given vectors are;
and 
a) To add or subtract two vectors, we add or subtract their corresponding components
b) See attachment for graphs(1,6
