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Vaselesa [24]
3 years ago
9

Mr. Rifkin had 240 digital and manual cameras. After donating 82 digital cameras and 26 manual cameras, he had three times as ma

ny digital cameras as manual cameras left. How many digital cameras did he have to begin with?
Mathematics
2 answers:
Rina8888 [55]3 years ago
4 0

Answer:

181

Step-by-step explanation:

puteri [66]3 years ago
3 0
After the donation,
3m + m = 132
manual = 33
digital = 99

Before the donation, he had 82 more digital and 26 more manual cameras.
82 + 99 = 181 digital
26 + 33 = 59 manual
For a double check, before donation,
digital + manual = 240
181 + 59 = 240
So, to begin with, the number of digital cameras = 181.


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Over 12 hours, it will travel 780 miles.

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Find the dimensions of the open rectangular box of maximum volume that can be made from a sheet of cardboard 21 in. by 12 in. by
a_sh-v [17]

Answer:

Dimension of the box is 16.1\times 7.1\times 2.45

The volume of the box is 280.05 in³.

Step-by-step explanation:          

Given : The open rectangular box of maximum volume that can be made from a sheet of cardboard 21 in. by 12 in. by cutting congruent squares from the corners and folding up the sides.

To find : The dimensions and the volume of the box?

Solution :

Let h be the height of the box which is the side length of a corner square.

According to question,

A sheet of cardboard 21 in. by 12 in. by cutting congruent squares from the corners and folding up the sides.

The length of the box is L=21-2h

The width of the box is W=12-2h

The volume of the box is V=L\times W\times H

V=(21-2h)\times (12-2h)\times h

V=(21-2h)\times (12h-2h^2)

V=252h-42h^2-24h^2+4h^3

V=4h^3-66h^2+252h

To maximize the volume we find derivative of volume and put it to zero.

V'=12h^2-132h+252

0=12h^2-132h+252

Solving by quadratic formula,

h=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

h=\frac{-(-132)\pm\sqrt{132^2-4(12)(252)}}{2(12)}

h=\frac{132\pm72.99}{24}

h=2.45,8.54

Now, substitute the value of h in the volume,

V=4h^3-66h^2+252h

When, h=2.45

V=4(2.45)^3-66(2.45)^2+252(2.45)

V\approx 280.05

When, h=8.54

V=4(8.54)^3-66(8.54)^2+252(8.54)

V\approx -170.06

Rejecting the negative volume as it is not possible.

Therefore, The volume of the box is 280.05 in³.

The dimension of the box is

The height of the box is h=2.45

The length of the box is L=21-2(2.45)=16.1

The width of the box is W=12-2(2.45)=7.1

So, Dimension of the box is 16.1\times 7.1\times 2.45

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