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ehidna [41]
2 years ago
8

Four circles, each with a radius of 2 inches, are removed from a square. What is the remaining area of the square?

Mathematics
1 answer:
garri49 [273]2 years ago
5 0

<u>Given</u>:

Given that the radius of the circle is 2 inches.

We need to determine the area of the remaining square.

<u>Area of a square:</u>

Given that each circle has a radius of 2 inches.

Then, the diameter of each circle is 4 inches.

Hence, the side length of the square is 2 × 4 = 8 inches.

The area of the square is given by

A=s^2

A=8^2

A=64 \ in^2

Thus, the area of the square is 64 square inches.

<u>Area of the four circles:</u>

The area of one circle is given by

A=\pi r^2

Substituting r = 2, we have;

A=4 \pi

Thus, the area of one circle is 4π in²

The area of 4 circles is 4 × 4π =16π in²

Hence, the area of the 4 circles is 16π in²

<u>Area of the remaining square:</u>

The area of the remaining square is given by

Area = Area of the square - Area of four circles.

Substituting the values, we get;

Area = 64-16 \pi

Thus, the area of the remaining square is (64 - 16π) in²

Hence, Option c is the correct answer.

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Answer:

The dimensions that minimize the surface are:

Wide: 1.65 yd

Long: 3.30 yd

Height: 2.20 yd

Step-by-step explanation:

We have a rectangular base, that its twice as long as it is wide.

It must hold 12 yd^3 of debris.

We have to minimize the surface area, subjet to the restriction of volume (12 yd^3).

The surface is equal to:

S=2(w*h+w*2w+2wh)=2(3wh+2w^2)

The volume restriction is:

V=w*2w*h=2w^2h=12\\\\h=\frac{6}{w^2}

If we replace h in the surface equation, we have:

S=2(3wh+2w^2)=6w(\frac{6}{w^2})+4w^2=36w^{-1}+4w^2

To optimize, we derive and equal to zero:

dS/dw=36(-1)w^{-2} + 8w=0\\\\36w^{-2}=8w\\\\w^3=36/8=4.5\\\\w=\sqrt[3]{4.5} =1.65

Then, the height h is:

h=6/w^2=6/(1.65^2)=6/2.7225=2.2

The dimensions that minimize the surface are:

Wide: 1.65 yd

Long: 3.30 yd

Height: 2.20 yd

8 0
2 years ago
The speed, s, of the current in a certain whirlpool is modeled by s=300/d, where d is the distance from the center of the whirlp
Ivan

Answer:

The answer is C. As you move closer to the center of the whirlpool, the speed of the current approaches infinity.

This means the closer you get, the faster you go, and the further the distance is the slower the speed. Hope this helps!

(Btw I got this right on the test)


6 0
2 years ago
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Agent Hunt is transferring classified files from the CIA mainframe to his flash drive. The variable S models the size of the fil
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Answer:

<h2>50MB</h2>

Step-by-step explanation:

\bold{METHOD\ 1:}\\\\\text{Calculate the value of }S\ \text{in}\ 10s,\ 20s,\ 30s,\ ... \text{and difference if them}\\\\for\ t=10:\\\\S=5(10)+45=50+45=95\\\\for\ t=20:\\\\S=5(20)+45=100+45=145\\\\for\ t=30:\\\\S=5(30)+45=150+45=195\\\\195-145=50\\145-95=50\\\\\boxed{50\ MB\ per\ 10s}

\bold{METHOD\ 2:}\\\\S_1=5t+45\to \text{in t second}\\\\S_2=5(t+10)+45\to \text{10 second leter}\\\\\text{Difference}\\\\S_2-S_1=5(t+10)+45-(5t+45)=5t+50+45-5t-45=50\\\\\boxed{50MB\ per\ 10s}

6 0
2 years ago
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Suppose that demand in period 1 was 7 units and the demand in period 2 was 9 units. Assume that the forecast for period 1 was fo
Zepler [3.9K]

Answer:

Step-by-step explanation:

Forecast for period 1 is 5

Demand For Period 1 is 7

Demand for Period  2 is 9  

Forecast  can be given by

F_{t+1}=F_t+\alpha (D_t-F_t)

where

F_{t+1}=Future Forecast

F_t=Present\ Period\ Forecast

D_t=Present\ Period\ Demand

\alpha =smoothing\ constant  

F_{t+1}=5+0.2(7-5)

F_{t+1}=5.4

Forecast for Period 3

F_{t+2}=F_{t+1}+\alpha (D_{t+1}-F_{t+1})

F_{t+2}=5.4+0.2\cdot (9-5.4)

F_{t+2}=6.12  

8 0
2 years ago
Kayleigh babysat for 11 hours this week. That was 5 fewer than 2/3 as many hours as she babysat last week, h. Write an equation
lukranit [14]

Answer: The answer is \textup{x}=\dfrac{2}{3}\textup{h}-5.


Step-by-step explanation: Given that Kayleigh babysat for 11 hours the present week. Also, this was 5 less than two-third of the number of hours she babysat last week, which is represented by 'h'.

We are to write an equation to represent the number of hours she babysat each week.

So, for that, let 'x' be the number of hours she babysat this week. Then, according to the question, we can write

\textup{x}=\dfrac{2}{3}\textup{h}-5.

Also, it is given that

\textup{x}=11.

Therefore,

11=\dfrac{2}{3}\textup{h}-5\\\\\Rightarrow \dfrac{2}{3}\textup{h}=16\\\\\Rightarrow \textup{h}=24.

Hence, using the above relation, we can find the number oh hours Keyleigh babysat each week.

Thus, the required equation is

\textup{x}=\dfrac{2}{3}\textup{h}-5,

where, 'x' and 'h' are the number of hours she sat this week and last week respectively.

 


4 0
2 years ago
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