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Xelga [282]
3 years ago
13

Can someone assist me with these Algreba questions olease?

Mathematics
1 answer:
Delicious77 [7]3 years ago
4 0

the answer have to be would be (-3, 1)

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2.268 nearest whole gram
MissTica

Answer:

It will be 2.27

Since the 8 in 2.268 is larger than 5, the 6 will turn to the next number. So, the answer will be 2.27.

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Simplify (3x-5)+(5x+1)
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Group the common terms together

3x + 5x - 5 + 1

then add or subtract them

8x - 4
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Can someone please answer this question please answer it correctly and please show work please help me I need it
Marina CMI [18]

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-9

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8 0
3 years ago
The ubiquitous 12oz aluminum cans used to distribute drinks in this country have a diameter of approximately 2.75 inches and a h
gayaneshka [121]

Answer:

3.5%

Step-by-step explanation:

The volume of a cylinder = \pi r^2h

<em>r</em> = radius of cylinder,

<em>h</em> = height of cylinder

For the non-optimal can,

<em>r</em> = 2.75/2 = 1.375

<em>h</em> = 5.0

V = \pi(1.375^2)\times 5.0 = 9.453125\pi

<em />

For the optimal can,

<em>d</em>/<em>h</em> = 1,

<em>d</em> = <em>h</em>

2<em>r </em>=<em> h</em>

<em>r</em> = h/2

V = \pi\left(\dfrac{h}{2}\right)^2\times h = \pi\left(\dfrac{h^3}{4}\right)

They have the same volume.

<em />\pi\dfrac{h^3}{4} = 9.453125\pi<em />

h^3 = 37.8125

h=3.36 (This is the height of the optimal can)

r = \dfrac{3.36}{2} = 1.68 (This is the radius of the optimal can)

The area of a cylinder is

<em />A = 2\pi r(r+h)<em />

For the non-optimal can,

A = 2\pi\times\dfrac{2.75}{2}\left(\dfrac{2.75}{2}+5.0\right) = 17.53125\pi

For the optimal can,

A = 2\pi\times1.68\left(1.68+3.36\right) = 16.9344\pi

Amount of aluminum saved, as a percentage of the amount used to make the optimal cans = \dfrac{17.53125\pi - 16.9344\pi}{16.9344\pi}\times 100\% = 3.5\%

5 0
3 years ago
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