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3 years ago

How many feet are in 333 miles?

Mathematics

1 answer:

zheka24 [161]3 years ago

6 0

Answer:

1,758,240 Feet

Step-by-step explanation:

1 Mile = 5280 Feet

333 Miles = 1,758,240 Feet

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Let U = {q, r, s, t, u, v, w, x, y, z} A = {q, s, u, w, y} B = {q, s, y, z} C = {v, w, x, y, z}. List the elements in the set. (

algol13

Answer:

its correct answer is

{y}

7 0

3 years ago

It took 30 minutes for Melinda to run 4 miles what is the ratio

ICE Princess25 [194]

7.5minutes/1miles
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5 0

4 years ago

Solve for X<br> (5x-2)^2=17<br> please help me everyone.

beks73 [17]

Answer:

$\frac{2}{5}±\frac{\sqrt{17}}{5}$

Step-by-step explanation:

$(5x-2)^2=17 \\ 5x-2=±\sqrt{17} \\ 5x = 2 ± \sqrt{17} \\ x=\frac{2}{5}± \frac{\sqrt{17}}{5}$

Don't mind the $±$

5 0

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

Can u please help me with this math assignment

Troyanec [42]

Hard to read sideways but what I got from that is that you have a right triangle and you are trying to explain the other angles?
Both of the other two angles would be 45 because an angle is 180 total. If it is a right angle, that means that there is a 90 degree angle. 180-90=90. In a right angle, the other two angles are the same in degree. Therefor 90/2 is 45. 45 applies to both of the acute angles. I hope this helps! Good luck :)

6 0

3 years ago