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melisa1 [442]
3 years ago
13

A snow cone cup is 12 cm tall and has a diameter of 8 cm. Find the volume of flavored ice that can be held inside the snow cone

cup.
Mathematics
2 answers:
Vinvika [58]3 years ago
8 0

Answer:

201.06cm ^3

Step-by-step explanation:

Marina86 [1]3 years ago
7 0
201.06cm^3is the 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.

8 0
3 years ago
<img src="https://tex.z-dn.net/?f=%5Csqrt%7Bx%7D%2064%2F4" id="TexFormula1" title="\sqrt{x} 64/4" alt="\sqrt{x} 64/4" align="abs
ozzi

Answer: =16\sqrt{x}

Step-by-step explanation:

\sqrt{x}\frac{64}{4}=16\sqrt{x}

8 0
2 years ago
In trapezoid KLMN, solve for x ,then use that value to solve for
Andre45 [30]

Answer:

<u>Same side angles are supplementary</u>

L+k=180

18x-18+72=180

18x=126

x=7

m∠18(7)-18

= 108°

---------------------

hope it helps...

have a grat day

3 0
3 years ago
Which are the solution of x2=-5x+8
yawa3891 [41]

rewrite the equation set = to 0.

x^2 + 5x - 8 = 0

The quadratic will not factor so you have to use the quadratic formula.

x = (-b + - sqrt(b^2 - 4ac))/2a

x = (5 + - sqrt(25 - 4* 1* -8))/2

x = (5 + - sqrt 57)/2


The x2is not the same as 2x. It is x^2. X tot he second power which makes the problem a quadratic equation. You cannot combine the terms x^2 and -5x because they so not have the same power.

4 0
3 years ago
25+ POINTS WILL MARK BRAINLIEST
Katen [24]

Answer:

Point A = (-6,0)

Point B = (0,4)

7 0
3 years ago
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