Algebra 2 Help-10 pts?

A = 2πr² + 2πrh
A - 2πr² = 2πrh
h = (A - 2πr²) / (2πr)

Answer: h = A/(2πr) - r

a = (pi)r² + 2(pi)rh
2(pi)rh = a - (pi)r²
h = (a - [pi]r²])/(2[pi]r)

Answer: (a - [pi]r²])/(2[pi]r)

first one, minus 0.9 from itself and from -one million.5 youll get 0.8x= ____ <-the variation of -one million.5-0.9 then divide by 0.8 from the two components to cancel out the 0.8 x= in spite of the variation of -one million.5-0.9 / 0.8 2d one.. 6(x) = 6x+ 6(one million.37)=5x .. 6x+7.37=5x minus 6 x from the two components 7.37=-x divide by -1x on the two components x=-7.37

A = 2πr^2 + 2πrh
2πrh = A - 2πr^2
h = (A - 2πr^2)/2πr
h = A/2πr - 2πr^2/2πr
h = A/2πr - r

A = 2 π r ² + 2 π r h
2 π r h = A - 2 π r ²
h = A / (2 π r) - r
OR
h = ( A - 2 π r ² ) / (2 π r)

but there are 2 unknowns, and only 1 equation

you mean the value for h? is a mean something?

2 pi r h = A - 2 pi r^2

so

h = (A- 2 pi r^2) / 2 pi r

h = (A / 2 pi r) - r

h = (A - 2 * pi * r^2) / 2 * pi * r

Just subtract the first term then divide both sides by 2 * pi * r. This will leave h by itself.

A = 2*pi*r^2 + 2*pi*rh

This looks remarkably like the surface area of a cylinder :-)

2*pi*r*h = A - 2*pi*r^2

h = (A - 2*pi*r^2) / (2*pi*r)

h = A / (2*pi*r) - r