All categories

2 years ago

Find the area of the figure shown in the diagram to the nearest hundredth. Help pliss

Mathematics

2 answers:

saw5 [17]2 years ago

4 0

Answer:

41.13m

Step-by-step explanation:

To find the area of a circle, use A= πr²

A(of circle) = π * 4²

A(of circle) = π * 16

A(of circle) ≈ 50.2655

Because you only have 2 quarters of a circle, you multiply it by 2/4

A(of 2/4 circle) ≈ 25.13275

A of rectangle = l * w

A(of rectangle) = 4 * 4

A(of rectangle) = 16

16+25.13275 = 41.13275

or 41.13

denis23 [38]2 years ago

3 0

Answer:

41.13 sq m

Step-by-step explanation:

area of square = 4² = 16

there are two quarter-circles which equals one semicircle

area of semi-circle is πr²/2

A = 4²π/2 = 8π = 25.13

add together: 16 + 25.13 = 41.13

You might be interested in

If m = 35 cm and n = 37 cm, what is the length of l?

ElenaW [278]

Minus m-n = 36 so 36 is your answer.

6 0

2 years ago

Read 2 more answers

What is square root of 167?

enot [183]

The square root of 167 would be 12.922847. If you need it rounded to the nearest tenth it would be 12.9

If you need t rounded to the nearest hundredth, it would be 12.92.

Hope this helps :)

7 0

2 years ago

What is a GPA? How is it calculated and determined? If a person is required to have a GPA 95 or higher, what does that mean?

tester [92]

Answer:

GPA stands for grade point average. the highest you can get is 4.5. There is no such things as a 95 GPA. To calculate, take the number or point per grade you have, A is 4.0, B is 3.0, C is 2.0, and so on, and add them all up and then divide by the number of classes you are taking.

Step-by-step explanation:

4 0

3 years ago

1-(1/3) x 1-(1/3) x 1/3

PolarNik [594]

The solution is -4/9x + 1

7 0

3 years ago

Read 2 more answers

PLS ANSWER ASAP 30 POINTS!!! CHECK PHOTO! WILL MARK BRAINLIEST TO WHO ANSWERS

Sveta_85 [38]

I'll do Problem 8 to get you started

a = 4 and c = 7 are the two given sides

Use these values in the pythagorean theorem to find side b

$a^2 + b^2 = c^2\\\\4^2 + b^2 = 7^2\\\\16 + b^2 = 49\\\\b^2 = 49 - 16\\\\b^2 = 33\\\\b = \sqrt{33}\\\\$

With respect to reference angle A, we have:

opposite side = a = 4
adjacent side = b = $\sqrt{33}$
hypotenuse = c = 7

Now let's compute the 6 trig ratios for the angle A.

We'll start with the sine ratio which is opposite over hypotenuse.

$\sin(\text{angle}) = \frac{\text{opposite}}{\text{hypotenuse}}\\\\\sin(A) = \frac{a}{c}\\\\\sin(A) = \frac{4}{7}\\\\$

Then cosine which is adjacent over hypotenuse

$\cos(\text{angle}) = \frac{\text{adjacent}}{\text{hypotenuse}}\\\\\cos(A) = \frac{b}{c}\\\\\cos(A) = \frac{\sqrt{33}}{7}\\\\$

Tangent is the ratio of opposite over adjacent

$\tan(\text{angle}) = \frac{\text{opposite}}{\text{adjacent}}\\\\\tan(A) = \frac{a}{b}\\\\\tan(A) = \frac{4}{\sqrt{33}}\\\\\tan(A) = \frac{4\sqrt{33}}{\sqrt{33}*\sqrt{33}}\\\\\tan(A) = \frac{4\sqrt{33}}{(\sqrt{33})^2}\\\\\tan(A) = \frac{4\sqrt{33}}{33}\\\\$

Rationalizing the denominator may be optional, so I would ask your teacher for clarification.

So far we've taken care of 3 trig functions. The remaining 3 are reciprocals of the ones mentioned so far.

cosecant, abbreviated as csc, is the reciprocal of sine
secant, abbreviated as sec, is the reciprocal of cosine
cotangent, abbreviated as cot, is the reciprocal of tangent

So we'll flip the fraction of each like so:

$\csc(\text{angle}) = \frac{\text{hypotenuse}}{\text{opposite}} \ \text{ ... reciprocal of sine}\\\\\csc(A) = \frac{c}{a}\\\\\csc(A) = \frac{7}{4}\\\\\sec(\text{angle}) = \frac{\text{hypotenuse}}{\text{adjacent}} \ \text{ ... reciprocal of cosine}\\\\\sec(A) = \frac{c}{b}\\\\\sec(A) = \frac{7}{\sqrt{33}} = \frac{7\sqrt{33}}{33}\\\\\cot(\text{angle}) = \frac{\text{adjacent}}{\text{opposite}} \ \text{ ... reciprocal of tangent}\\\\\cot(A) = \frac{b}{a}\\\\\cot(A) = \frac{\sqrt{33}}{4}\\\\$

------------------------------------------------------

Summary:

The missing side is $b = \sqrt{33}$

The 6 trig functions have these results

$\sin(A) = \frac{4}{7}\\\\\cos(A) = \frac{\sqrt{33}}{7}\\\\\tan(A) = \frac{4}{\sqrt{33}} = \frac{4\sqrt{33}}{33}\\\\\csc(A) = \frac{7}{4}\\\\\sec(A) = \frac{7}{\sqrt{33}} = \frac{7\sqrt{33}}{33}\\\\\cot(A) = \frac{\sqrt{33}}{4}\\\\$

Rationalizing the denominator may be optional, but I would ask your teacher to be sure.

7 0

1 year ago