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TEA [102]
3 years ago
5

Zeus Industries bought a computer for $2207. It is expected to depreciate at a rate of 27% per year. What will the value of the

computer be in 4 years?
Mathematics
1 answer:
vodka [1.7K]3 years ago
7 0
Multiply 2207 by .27 and get $595.89.  This is the first year depreciation.
After 1 year the computer has a value of $1611.11.
Year 2 depreciation would be 1611.11 X .27= $434
1611.11-434= $1,177.11
Year 3  1177.11 X .27= $317.82
1177.11- 317.82= $859.29
Final year 859.29 X .27= 232
$859.29- 232= $627.29
You might be interested in
The perimeter of a rectangular piece of plastic is 30 millimeters. The area is 54 square millimeters. What are the dimensions of
Delvig [45]

Answer:

6 mm and 9 mm are the dimensions of the piece of plastic.

Step-by-step explanation:

Keep in mind the formulas for the area and perimeter of a rectangle:

A = lw

P = 2 (l + w)

List the factors of 54:

1, 2, 3, 6, 9, 18, 27, 54

POSSIBLE DIMENSIONS of the piece of plastic:

1 mm and 54 mm:

Area - 54 mm^2

Perimeter - 110 mm

2 mm and 27 mm

Area - 54 mm^2

Perimeter - 58 mm

3 mm and 18 mm

Area - 54 mm^2

Perimeter - 42 mm

6 mm and 9 mm

Area - 54 mm^2

Perimeter - 30 mm

The rectangular piece of plastic with the dimensions 6mm and 9 mm corresponds with the area and perimeter of the piece of plastic mentioned. So these are the correct dimensions.

Hope this helps!

6 0
2 years ago
Will give brainliest.
Dennis_Churaev [7]
Pretty sure it's A. (X times .5 equals Y)
8 0
2 years ago
Find the area of the sector with a central angle of 120° and a radius of 8 inches. Leave in terms of π
soldier1979 [14.2K]
Centralangle/360 times area of circle=sector area


120/360 times pi8²=
(1/3)(64pi)=64pi/3 square inches
8 0
3 years ago
Read 2 more answers
Please help if you can!
Dima020 [189]

Answer:

m' (-2, 4)

n' (-2, 1)

o' (3, 1)

p' (3, 4)

Step-by-step explanation:

4 0
2 years ago
Read 2 more answers
Find the surface area of the part of the sphere x2+y2+z2=81 that lies above the cone z=x2+y2−−−−−−√
TiliK225 [7]
Parameterize the part of the sphere by

\mathbf s(u,v)=(9\cos u\sin v,9\sin u\sin v,9\cos v)

with 0\le u\le2\pi and 0\le v\le\dfrac\pi4. Then

\mathrm dS=\|\mathbf s_u\times\mathbf s_v\|=81\sin v\,\mathrm du\,\mathrm dv

So the area of S (the part of the sphere above the cone) is given by

\displaystyle\iint_S\mathrm dS=81\int_{v=0}^{v=\pi/4}\int_{u=0}^{u=2\pi}\sin v\,\mathrm du\,\mathrm dv=81(2\pi)\left(1-\dfrac1{\sqrt2}\right)=(162-81\sqrt2)\pi
4 0
3 years ago
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