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3 years ago

What is the approximate area of the figure? Help pls.....

Mathematics

1 answer:

Vaselesa [24]3 years ago

7 0

Okay, first we need to calculate the area of the rectangle.

Area of Rectangle- 16 x 6 = 96

Next, we need to calculate the area of the semicircles.

Formula: Radius x Radius x pi (3.14) = Area of circle

Since we know that the diameter is 6 we can figure out the radius.
The radius is half of the diameter. So our radius is 3.

3 x 3 x 3.14 = 28.26

Since it's a semi circle we now need to divide by 2.

28.26 divided by 2 = 14.13

Then, we add.

Area of rectangle
+
Area of 2 semi circles

96 + 14.13 + 14.13 = Total Area of The Figure

So our answer is 124.26

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Find the missing side x

bezimeni [28]

Answer:

$\sqrt{968}$

Step-by-step explanation:

Since this is a right triangle, we are able to use pythagorean theorem, a^2+b^2=c^2. In this case x would be the "c", so 22^2+22^2=x^2. Isolate the variable and solve for x. 484+484=x^2

968=x^2

$\sqrt{968\\}$ =x

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3 years ago

Find the quotient and explain the process and reasoning. (PLEASE EXPLAIN IT MATHEMATICALLY, like in numbers!)

patriot [66]

9 because 27 divided by 3 is 9 and because both numbers don’t have a whole number it equals a whole number(sorry if I didn’t explain good enough)

8 0

1 year ago

Which values are in the solution set of the compound inequality? Check all that apply. 4(x + 3) ≤ 0 or x + 1 > 3 –6 –3 0 3 8

Colt1911 [192]

we have

$4(x + 3) \leq 0$ -------> inequality 1

$x + 1 > 3$ -------> inequality 2

we know that

In this system of inequalities, for a value to be the solution of the system, it is enough that it satisfies at least one of the two inequalities.

let's check each of the values

case 1) x=-6

Substitute the value of x=-6 in the inequality 1

$4(-6 + 3) \leq 0$

$4(-3) \leq 0$

$-12 \leq 0$ -------> is ok

The value of x=-6 is a solution of the compound inequality-----> It is not necessary to check the second inequality, because the first one satisfies

case 2) x=-3

Substitute the value of x=-3 in the inequality 1

$4(-3 + 3) \leq 0$

$4(0) \leq 0$

$0 \leq 0$ -------> is ok

The value of x=-3 is a solution of the compound inequality-----> It is not necessary to check the second inequality, because the first one satisfies

case 3) x=0

Substitute the value of x=0 in the inequality 1

$4(0 + 3) \leq 0$