Abowl in the shape of a hemispere is filled with water to a depth h=3 inches. the radius of the bowl is r inches. express the radius of the bowl r as a function of the angle theta.

The radius of the bowl R as a function of the angle theta is

Solution:

The figure is attached below

If we consider the centre of hemisphere be A

The radius be AC and AD

According to question,  

A bowl in the shape of a hemispere is filled with water to a depth h=3 inches .i.e. BC = h = 3 inches

And radius of the bowl is R inches .i.e. R = AD =AC

Now , using  trigonometric identities in triangle ABD we get

Since , AC = AB + BC

R = R Sinθ + 3

R - R Sinθ = 3

R (1 – Sinθ ) = 3

Which is the required expression for the radius of the bowl R as a function of the angle theta

the answer is 844

step-by-step explanation:

it is 844 because if you add 836.50 + 7.5 you will get 844

the smallest possible measurement of the width is 14 feet

step-by-step explanation:

let's assume length of parallelogram is 'l'

width of parallelogram is 'w'

the perimeter of a parallelogram must be no less than 40 feet

so, we get

we are given

length of parallelogram is 6 feet

so,

l=6

we can plug it

now, we can solve for w

subtract both sides by 12

divide both sides by 2

and we get

so,

the smallest possible measurement of the width is 14 feet