# What is the area of this figure? :)

Answer: there are 4 semicircles, each radius is 8cm, so we get the total areas of these semicircles: 4×0.5×π
and the square area: 16²
so the total area of figure: 16²+4×0.5×π=256+128π

## Related Questions

Pls help due tomorrow just numbers 7 - 12

Step-by-step explanation:

7). 5.75% = = =

10). % = % = =

12). % = % =

Now is you turn. You can do it!

I need help with math ​

The sixth grade class is ordering special t-shirts. There are 88 sixth graders in all. The only sizes they can select are small, medium, or large. The class needs five times as many medium shirts as small shirts, and twice as many large shirts as small shirts. Let s represent the number of small shirts. Use any method (algebraically, tables, or tape diagrams) to show your work and solve for how many small, medium, and large shirts the class will need to order.

166,000

Step-by-step explanation:

1x166,000=166,000

Hello I am needing help on this word problem please. I think it's 30 but I'm questioning myself

Solution:

Since 5 pounds of meat will feed about 26 people.

Thus;

If she is expecting 156 people, she should prepare;

An object is launched at 20 m/s from a height of 65 m. The equation for the height (h) in terms of time (t) is given by h(t) = -4.912 + 20t + 65. What is theobject's maximum height?

The maximum height is the y-value of the vertex.

h(t) = -4.9t² + 20t + 65

a=-4.9   b=20  c=65

h(2) = -4.9(2)² + 20(2) + 65

= -19.6 + 40 + 65

= 85.4

Step-by-step explanation:

The weights, in pounds, of the cats in an animal adoption center are normally distributed. If a random sample of cats is taken and the confidence interval is (7.9,12.7), what is the margin of error

Step-by-step explanation:

Given : The weights, in pounds, of the cats in an animal adoption center are normally distributed.

The confidence interval is (7.9,12.7)           (1)

Let be the sample mean of the weights, in pounds, of the cats in an animal adoption center.

We know that the confidence interval for population mean is given by :-

(2)

From (1) and (2)

Subtracting (3) from (4), we get

Hence, the margin of error is E = 2.4