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3 years ago

I need help on this

Mathematics

1 answer:

Goshia [24]3 years ago

6 0

I'm pretty sure the answer is A

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One week, Kerry travels 125 miles and uses 5 gallons of gasoline. The next week, she travels 175 miles and uses 7 gallons of gas

bazaltina [42]

This is good information, but the question part is missing. Put the question in the comments and I'll try my best to help!

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3 years ago

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 $L\approx 15.57$ .

8 0

3 years ago

If a=under root s(s-a)(s-b)(s-c) then the value of a, when a =3 b =4c = 5 and s =a+b+c by 2

kramer

Answer:

Step-by-step explanation:

Here it's given that ,

$\sf\red{\longrightarrow} A=\sqrt{s(s-a)(s-b)(s-c)}$

Also ,

$\sf\red{\longrightarrow} s =\dfrac{a+b+c}{2}$

And we need to find out the value of A , when

a = 3
b = 4
c = 5

So , on substituting the respective values to find s we have ,

$\sf\red{\longrightarrow} s =\dfrac{3+4+5}{2}=\dfrac{12}{2}=\bf{6}$

Now let's find out A as ,

$\sf\red{\longrightarrow} A = \sqrt{6(6-3)(6-4)(6-5)} \\$

$\sf\red{\longrightarrow}A =\sqrt{6 * 3 * 2 * 1}\\$

$\sf\red{\longrightarrow} A = \sqrt{3^2 * 2^2 *1^2}\\$

$\sf\red{\longrightarrow}A = 3*2\\$

$\sf\red{\longrightarrow}\boxed{\qquad \blue{\bf A = 6 \qquad}}$

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2 years ago

What does it mean when a researcher states that an observed effect or difference is significant with a confidence level of 99%?

icang [17]

Answer:

it affected more than half

Step-by-step explanation:

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3 years ago

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Oliver made 8 pints of jam last year. How many pints does he make this year?

Gekata [30.6K]

Answer:

23 pints

Step-by-step explanation:

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3 years ago

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