Use differentials to estimate the amount of material in a closed cylindrical can that is 50 cm high and 20 cm in diameter if the metal in the top and bottom is 0.1 cm thick, and the metal in the sides is 0.05 cm thick. Note, you are approximating the volume of metal which makes up the can (i. e. melt the can into a blob and measure its volume), not the volume it encloses. The differential for the volume is

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step-by-step explanation:

10000

step-by-step explanation: divide by 2,

what are the statments

step-by-step explanation:

g(x) = 3*(8(3^(x−2))+2)

g(x) = 24 (3^(x-2) )+6

the factor is 2

step-by-step explanation:

let f(x)=8(3^(x−2)+2)

to stretch a graph by 3, it would be 3f(x)

g(x) =3f(x)

g(x) = 3*(8(3^(x−2))+2)

g(x) = 24 (3^(x-2) )+6

x     −2     0     2     4     6

f(x) 1/64 1/16   1/4     1     4

1/64 * z * z = 1/16

i multiplies by z twice because it its 2 steps from -2 to 0

1/64 z^2 = 1/16

multiply by 64 to clear the fraction on the left side

1/4 * 64 z^2 =1/16 *64

z^2 = 4

take the square root of each side

sqrt(z^2) = ±sqrt(4)

z = ±2  

we are multiplying by 2 each time.