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Andre45 [30]
3 years ago
6

I am having trouble with this ellipse problem. Please help right away​

Mathematics
1 answer:
Vitek1552 [10]3 years ago
3 0

Answer:

a. The distance from the center to either vertex

Step-by-step explanation:

The distance from the center to a vertex is the fixed value <em>a</em>. The values of <em>a</em> and <em>c</em> will vary from one ellipse to another, but they are fixed for any given ellipse.

I hope this helps you out alot, and as always, I am joyous to assist anyone at any time.

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-7

Step-by-step explanation:

Refer to the photo.

the slope=(0-g)/(5-4)

i.e. 7=(-g)/1

g=-7

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3 years ago
A = πηγ2<br> What do you divide by?
defon

Answer:

76.20+q; 76.60Es

Step-by-step explanation:

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2 years ago
If the price of a business math text rose to $120 and this was 10% more than the original price what was the original selling pr
kolezko [41]

Step-by-step explanation:

120*10% =12

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= (10/100)*120

= (10*120)/100

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5 0
2 years ago
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4 0
2 years ago
An open top box is to be built with a rectangular base whose length is twice its width and with a volume of 36 ft 3 . Find the d
denpristay [2]

Answer:

The dimensions of the box that minimize the materials used is 6\times 3\times 2\ ft

Step-by-step explanation:

Given : An open top box is to be built with a rectangular base whose length is twice its width and with a volume of 36 ft³.

To find : The dimensions of the box that minimize the materials used ?

Solution :

An open top box is to be built with a rectangular base whose length is twice its width.

Here, width = w

Length = 2w

Height = h

The volume of the box V=36 ft³

i.e. w\times 2w\times h=36

h=\frac{18}{w^2}

The equation form when top is open,

f(w)=2w^2+2wh+2(2w)h

Substitute the value of h,

f(w)=2w^2+2w(\frac{18}{w^2})+2(2w)(\frac{18}{w^2})

f(w)=2w^2+\frac{36}{w}+\frac{72}{w}

f(w)=2w^2+\frac{108}{w}

Derivate w.r.t 'w',

f'(w)=4w-\frac{108}{w^2}

For critical point put it to zero,

4w-\frac{108}{w^2}=0

4w=\frac{108}{w^2}

w^3=27

w^3=3^3

w=3

Derivate the function again w.r.t 'w',

f''(w)=4+\frac{216}{w^3}

For w=3, f''(3)=4+\frac{216}{3^3}=12 >0

So, it is minimum at w=3.

Now, the dimensions of the box is

Width = 3 ft.

Length = 2(3)= 6 ft

Height = \frac{18}{3^2}=2\ ft

Therefore, the dimensions of the box that minimize the materials used is 6\times 3\times 2\ ft

4 0
3 years ago
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