Somebody please help me with 10+9+(9-7)*5

6 feet 12 feet What is the area of the given composite shape?

Reika [66]

Answer:

choice A; 128.52 square feet

Step-by-step explanation:

This shape is made up of a semi-circle and a rectangle. The area of the rectangle will be 12*6 = 72 sq ft.

The formula for area of a circle is $\pi$ * $r^{2}$ . also r=radius. We are given the diameter, 12 feet. The radius is diameter/2, so the radius of the circle will be 6 feet.

Since pi is infinite we round it to 3.14.

Plug in the values and our expression will be (3.14* $6^{2}$ )/2 because it is a semi-circle. Simplify that to (3.14*36)/2 = 113.04/2 = 56.52 sq ft.

Remember to add the area of the rectangle and semi-circle. 72 + 56.52 = 128.52 sq ft.

3 0

2 years ago

James Wants to triple his vegetarian meatball recipe that uses 1 lb. 10 oz. of tofu how much tofu will he need in total

AVprozaik [17]

Answer: The tripled tofu is 4 lb. 14 oz. Explanation: So you first want to turn your mass into the same unit, so change the pounds of tofu into ounces. There are 16 ounces in a pound, which means the total mass of the original tofu could be written as (16 oz.) + 10 oz. = 26 oz. tofu. Multiply this by 3, and you have tripled the tofu. 3 * 26 = 78. To simplify it, divide 78 oz by 16 to find pounds, and leave the remainder in ounces. 78/16 = 4 lb. 14 oz tofu.

7 0

3 years ago

Find thhe remainder when 7^203 is divided by 4

Aleksandr [31]

Using the square-and-multiply approach, we have

$7^{203}=7\times(7^{101})^2$
$7^{101}=7\times(7^{50})^2$
$7^{50}=(7^{25})^2$
$7^{25}=7\times(7^{12})^2$
$7^{12}=(7^6)^2$
$7^6=(7^3)^2$
$7^3=7\times7^2$

and so, using the property that, if $a_1\equiv b_1\mod n$ and $a_2\equiv b_2\mod n$ , then $a_1a_2\equiv b_1b_2\mod n$ , we get

$7\equiv3\mod4$
$7^2\equiv9\equiv1\mod4$
$7^3\equiv7\times1\equiv7\equiv3\mod4$
$7^6\equiv9\equiv1\mod4$
$7^{12}\equiv1\mod4$
$7^{25}\equiv7\times1\equiv7\equiv3\mod4$
$7^{50}\equiv9\equiv1\mod4$
$7^{101}\equiv7\times1\equiv7\equiv3\mod4$
$7^{203}\equiv7\times9\equiv3\times1\equiv3\mod4$