Use differentials to estimate the amount of tin in a closed tin can with diameter 12 cm and height 11 cm if the tin is 0.04 cm thick.

16.65 cm³

Step-by-step explanation:

Given that :

the diameter of the closed tin can = 12 cm

the height of the closed tin can = 11 cm

The thickness of the closed tin can  = 0.04 cm

The main task required to be carried out is to use differentials to estimate the amount of tin in a closed tin can.

We all know that the volume of a tin can =

The differential of the volume   is :

Differentiating V  with respect to r; we have:

Differentiating V with respect to h ; we have:

Hence; the estimate of the amount of tin in a closed tin can is:

=

=

=

where;

dr = increase in radius = 0.04cm (since thickness of the radius = 0.04 cm)

dh = increase in the height = 0.04 + 0.04 = 0.08

= 16.65 cm³

Hence; the amount of tin in a closed tin can  is = 16.65 cm³

first option

a function is a relation in which an input is paired with exactly one output.

so, let's find the right answer.

in our first option, our inputs are january, february, and march. the outputs are 30, 28, and 25. each input has it's own output. this is correct!

our second option we know is incorrect, because ms. valverade has two outputs, dogs and cats. this is not a function.

our third option is incorrect as well. the yellow square has two ouputs, the square and the triangle.

lastly, in our fourth option, we see the number 7 two times with a different output. therefore, this is not a function.

remember: each input must be paired with only one output!

look at the picture.

y-intercept for the function on a picture is 4 → point (0, 4).

y-intercept for the function

is for x = 0. substitute:

y-intercept is 8 → point (0, 8).