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Leya [2.2K]
3 years ago
14

Which is another way to write 42+63?

Mathematics
1 answer:
eduard3 years ago
7 0
Forty two plus sixty three that's your answer 
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Circle O has a circumference of approximately 250π ft
liraira [26]

Answer:

79.617...

Step-by-step explanation:

When you are trying to find the circumfrence you do c=πd

So to find the diameter you reverse it

Since you already know the 250 feet you divide it by π

You then get 79.6

5 0
3 years ago
True or False? A right triangle can have a leg measuring 10cm and a hypotenuse measuring 10cm
Citrus2011 [14]
True, and it's called an isosceles right triangle :D
5 0
3 years ago
Tonka weighs 150 lb. He is loading a freight elevator with identical 67-pound boxes. The elevator can carry no more than 2000 lb
tia_tia [17]

all

150+67=217 woohoo it's 217 not 2000 sooo him and all boxes

7 0
3 years ago
If 3:y=18:24 find y.
VashaNatasha [74]

y = 4

Explanation:\begin{gathered} 3\text{ : y = 18 : 24} \\ 3\colon y\text{ = }\frac{3}{y} \\ 18\colon24\text{ = }\frac{18}{24} \end{gathered}\begin{gathered} \frac{3}{y}\text{ = }\frac{18}{24} \\ \text{cross mult}iply\colon \\ 3(24)\text{ = y(18)} \\ \text{divide both sides by 18:} \\ \frac{3(24)}{18}\text{ = }\frac{18y}{18} \end{gathered}\begin{gathered} y\text{ = }\frac{24}{6} \\ y\text{ = 4} \end{gathered}

7 0
1 year 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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