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Nimfa-mama [501]
3 years ago
15

Select all that apply.Several pennies are tossed in the air. Twelve land heads up and fifteen land tails up. Which of the follow

ing ratios are correct?
12 heads to 27 coins

15 tails to 12 heads

5 tails to 9 coins

12 heads to 15 tails
Mathematics
2 answers:
OlgaM077 [116]3 years ago
7 0
The answer is the last one: twelve heads to fifteen tails. After all, those are the two numbers mentioned, right?
OverLord2011 [107]3 years ago
3 0
The information we've been given is:

12 heads
15 tails

From those, we find that there are 12 + 15 = 27 coins.

Let's examine these ratios as such:

- 12 heads to 27 coins -
There are indeed 12 heads and 27 coins, so this comparison agrees with our data.

- 15 tails to 12 heads - 
Again, nothing here contradicts with our data, so this would be correct as well

- 12 heads to 15 tails -
Simply a reversed version of the ratio above; still a completely valid way to compare the relative quantities of heads and tails

- 5 tails to 9 coins -
This one needs a little more examination. We observe that the ratio of tails to coins with our given data is 15 tails to 27 coins. This seems to go against our data, but we can simplify our ratio by dividing both the number of tails and the number of coins by 3, which indeed gives us the ratio 5 tails to 9 coins.

So, all of the given ratios are correct.
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Answer:


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A norman window is constructed by adjoining a semicircle to the top of an ordinary rectangular. Find the dimensions of a norman
Yanka [14]

Answer:

W\approx 8.72 and L\approx 15.57.

Step-by-step explanation:

Please find the attachment.

We have been given that a norman window is constructed by adjoining a semicircle to the top of an ordinary rectangular. The total perimeter is 38 feet.

The perimeter of the window will be equal to three sides of rectangle plus half the perimeter of circle. We can represent our given information in an equation as:

2L+W+\frac{1}{2}(2\pi r)=38

We can see that diameter of semicircle is W. We know that diameter is twice the radius, so we will get:

2L+W+\frac{1}{2}(2r\pi)=38

2L+W+\frac{\pi}{2}W=38

Let us find area of window equation as:

\text{Area}=W\cdot L+\frac{1}{2}(\pi r^2)

\text{Area}=W\cdot L+\frac{1}{2}(\pi (\frac{W}{2})^2)

\text{Area}=W\cdot L+\frac{\pi}{2}(\frac{W}{2})^2)

\text{Area}=W\cdot L+\frac{\pi}{2}(\frac{W^2}{4})

\text{Area}=W\cdot L+\frac{\pi}{8}W^2

Now, we will solve for L is terms W from perimeter equation as:

L=38-(W+\frac{\pi }{2}W)

Substitute this value in area equation:

A=W\cdot (38-W-\frac{\pi }{2}W)+\frac{\pi}{8}W^2

Since we need the area of window to maximize, so we need to optimize area equation.

A=W\cdot (38-W-\frac{\pi }{2}W)+\frac{\pi}{8}W^2  

A=38W-W^2-\frac{\pi }{2}W^2+\frac{\pi}{8}W^2  

Let us find derivative of area equation as:

A'=38-2W-\frac{2\pi }{2}W+\frac{2\pi}{8}W  

A'=38-2W-\pi W+\frac{\pi}{4}W    

A'=38-2W-\frac{4\pi W}{4}+\frac{\pi}{4}W

A'=38-2W-\frac{3\pi W}{4}

To find maxima, we will equate first derivative equal to 0 as:

38-2W-\frac{3\pi W}{4}=0

-2W-\frac{3\pi W}{4}=-38

\frac{-8W-3\pi W}{4}=-38

\frac{-8W-3\pi W}{4}*4=-38*4

-8W-3\pi W=-152

8W+3\pi W=152

W(8+3\pi)=152

W=\frac{152}{8+3\pi}

W=8.723210

W\approx 8.72

Upon substituting W=8.723210 in equation L=38-(W+\frac{\pi }{2}W), we will get:

L=38-(8.723210+\frac{\pi }{2}8.723210)

L=38-(8.723210+\frac{8.723210\pi }{2})

L=38-(8.723210+\frac{27.40477245}{2})

L=38-(8.723210+13.70238622)

L=38-(22.42559622)

L=15.57440378

L\approx 15.57

Therefore, the dimensions of the window that will maximize the area would be W\approx 8.72 and L\approx 15.57.

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I need help!!!!!!!!!!!!!!!
jarptica [38.1K]

Answer:

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Step-by-step explanation:

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