Please help me with this question thanks

Find the area of the shaded region. Round your answer to the nearest tenth. PLZ HELP

Ksju [112]

Answer:

30.96 m²

Step-by-step explanation:

In white there are four parts of a circle, so we can split the 12 m in half to show the radius, and then solve to find the area of the full circle.

A = r²

a = (6)²

a = (36)

a = 113.04

Then you find the area of the whole box.

12 x 12 = 144

Then you subtract the area of the circle in white from the area of the whole box.

144 - 113.04 = 30.96 m²

So that’s the area of the shaded region.

6 0

3 years ago

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What are your guys favorite marsupials?

IgorC [24]

Answer:

koalas

Step - by - step explanation:

because they're so cute and cuddly

3 0

2 years ago

Find the solution: -9(2x - 1) = 45 x = -4 x =23/-9 x = -2 x = 3

Aleks [24]

Answer:

x=-2

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Math question

strojnjashka [21]

Answer:

The candle has a radius of 8 centimeters and 16 centimeters and uses an amount of approximately 1206.372 square centimeters.

Step-by-step explanation:

The volume ( $V$ ), in cubic centimeters, and surface area ( $A_{s}$ ), in square centimeters, formulas for the candle are described below:

$V = \pi\cdot r^{2}\cdot h$ (1)

$A_{s} = 2\pi\cdot r^{2} + 2\pi\cdot r \cdot h$ (2)

Where:

$r$ - Radius, in centimeters.

$h$ - Height, in centimeters.

By (1) we have an expression of the height in terms of the volume and the radius of the candle:

$h = \frac{V}{\pi\cdot r^{2}}$

By substitution in (2) we get the following formula:

$A_{s} = 2\pi \cdot r^{2} + 2\pi\cdot r\cdot \left(\frac{V}{\pi\cdot r^{2}} \right)$

$A_{s} = 2\pi \cdot r^{2} +\frac{2\cdot V}{r}$

Then, we derive the formulas for the First and Second Derivative Tests:

First Derivative Test

$4\pi\cdot r -\frac{2\cdot V}{r^{2}} = 0$

$4\pi\cdot r^{3} - 2\cdot V = 0$

$2\pi\cdot r^{3} = V$

$r = \sqrt[3]{\frac{V}{2\pi} }$

There is just one result, since volume is a positive variable.

Second Derivative Test

$A_{s}'' = 4\pi + \frac{4\cdot V}{r^{3}}$

If $\left(r = \sqrt[3]{\frac{V}{2\pi}}\right)$ :

$A_{s} = 4\pi + \frac{4\cdot V}{\frac{V}{2\pi} }$

$A_{s} = 12\pi$ (which means that the critical value leads to a minimum)

If we know that $V = 3217\,cm^{3}$ , then the dimensions for the minimum amount of plastic are:

$r = \sqrt[3]{\frac{V}{2\pi} }$

$r = \sqrt[3]{\frac{3217\,cm^{3}}{2\pi}}$

$r = 8\,cm$

$h = \frac{V}{\pi\cdot r^{2}}$

$h = \frac{3217\,cm^{3}}{\pi\cdot (8\,cm)^{2}}$

$h = 16\,cm$

And the amount of plastic needed to cover the outside of the candle for packaging is:

$A_{s} = 2\pi\cdot r^{2} + 2\pi\cdot r \cdot h$

$A_{s} = 2\pi\cdot (8\,cm)^{2} + 2\pi\cdot (8\,cm)\cdot (16\,cm)$

$A_{s} \approx 1206.372\,cm^{2}$

The candle has a radius of 8 centimeters and 16 centimeters and uses an amount of approximately 1206.372 square centimeters.

3 0

3 years ago

(3m±4/5n)2 solve this question

sukhopar [10]

Answer:

(3m-4/5)2

Final result :

(15m - 4)2

——————————

52

Step by step solution :

Step 1 :

4

Simplify —

5

Equation at the end of step 1 :

4

(3m - —)2

5

Step 2 :

Rewriting the whole as an Equivalent Fraction :

2.1 Subtracting a fraction from a whole

Rewrite the whole as a fraction using 5 as the denominator :

3m 3m • 5

3m = —— = ——————

1 5

Equivalent fraction : The fraction thus generated looks different but has the same value as the whole

Common denominator : The equivalent fraction and the other fraction involved in the calculation share the same denominator

Adding fractions that have a common denominator :

2.2 Adding up the two equivalent fractions

Add the two equivalent fractions which now have a common denominator

Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:

3m • 5 - (4) 15m - 4

———————————— = ———————

5 5

Equation at the end of step 2 :

(15m - 4)

(—————————)2

5

Step 3 :

Final result :

(15m - 4)2

———

52

Step-by-step explanation:

5 0

2 years ago