Answer:
-(33+110z)/10z
Step-by-step explanation:
(-7/2z +4)+(1/5z -15)
from BODMAS, solving the one in the bracket
First(-7/2z +4)+(1/5z -15)
Find the Lcm
[-7/2z +4] +[1/5z -15]
(7+8z)/2z +(1-75z)/5z
find the Lcm which is 10z
[5(-7+8z) + 2(1-75z)]/10z
[(-35+40z)+(2-150z)]10z
(-35+2+40z-150z)/10z
(-33-110z)/10z
-(33+110z)/10z
Answer:
I did not understand your question.
But the answer is 40.0625
Answer:
x=15, y=5
Step-by-step explanation:
x+y=20
x-y=10
Adding both equations;
(x+x) + (y-y) = 20+10
2x = 30
x = 30/2 = 15
Substitute x=15 into x+y=20
y= 20-x = 20-15= 5

so notice above, the circle is centered at h,k or 0,0, the origin, and has a ratio of 5.
and y = 2, is just a horizontal line, check the picture below.
The candle has the shape of a cylinder with height H=6 in, and radius of the circular base R= (2.8)/2 =1.4 (in).
The volume of the cylinder is Area(base)*height =

(in cubed)
According to Cavalieri's principle, the volume of the candle is equal to the volume of the cylinder we described.
Answer: 37 in cubed