AB= 12m, BC= 16 cm and AD= 13 m.
Find the area of the shaded region.

100π - (96 + (13/2)√231 ) m^2

Step-by-step explanation:

We will solve this by finding the area of the triangles (the quadrilateral is cut by the diameter) and subtract it from the area of the circle.

There is a circle theorem which states that if a triangle is in the semicircle and the hypothenuse extends to the length of the diameter, the angle at B here is 90 degrees.

Since angle B is 90 degrees we can use the Pythagorean theorem (a^2 + b^2 = c^2) to find Line AC.

so 144 + 256 = 400

√400 = 20

So diameter is 20

Finding the area of the top triangle:

1/2(12x16) = 96cm^2

For the bottom triangle, we need the side DC to find the area. To find it we will apply the theorem once again.

400 = 169 + DC^2

DC^2 = 231

DC = √231

so the area of the bottom triangle is 1/2(13√231) = (13/2)√231

Now we add the area of the triangles for the quadrilateral and subtract from the circle area.

The circle area is 100π (πr^2)

Hence, the area of the shaded region is:

100π - (96 + (13/2)√231 ) m^2

step-by-step explanation:

ok one would be money, and the other would be refreshments or something similar. if each bag of popcorn is 5.00 and each drink is half of that that means drinks are 2.50 the proper way to do that would to look and see how much of each you can buy without exceeding a limit of 30.00

x =-.20752

step-by-step explanation:

4^(x+2)=12

4^(x+2)=12

take the log base 4 on each side

log 4 (4^(x+2) = log 4(12)

we know log b(a^y) = y log b (a)

(x+2) log 4(4) = log 4 (12)

x+2 = log 4 (12)

we know   logb c = loga c/loga b

we want to convert to base 10   so a = 10

by default, we do not write the 10   c = 4   and b = 10

log4 10 = log 12/log 4

log4 10 = log 12/ log 4

x+2 = log 12/ log 4

subtract 2 from each side

x +2 -2 = log 12/ log 4 -2

x = 1.79248125 -2

x =-.20752