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lesantik [10]
3 years ago
8

When originally purchased, a vehicle costing $25,200 had an estimated useful life of 8 years and an estimated salvage value of $

2,800. after 4 years of straight-line depreciation, the asset's total estimated useful life was revised from 8 years to 6 years and there was no change in the estimated salvage value. the depreciation expense in year 5 equals:?

Mathematics
1 answer:
hichkok12 [17]3 years ago
7 0
The depreciated value in year 4 is 14,000. In year 5, it is 8,400. The depreciation expense in year 5 is
.. $14,000 -8,400 = $5,600

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Can you PLEASE PLEASE help me with this question!!!!
svlad2 [7]
4% of 200 7 graders= 8, so 8 are expected to move by the end of the year. but if 12 students actually moved instead, there was 4 more moves than we expected. I hope this helped..!
4 0
3 years ago
Use the remainder theorem to find P (3) for P(x) = 2x² - 4x^2- 3.
ladessa [460]

Answer:

P(2) = 4

Step-by-step explanation:

P(X) = -2X4 + 4X3 - X + 6

Use the remainder theorem to find quotient and remainder and the value of P(2)

 

First add in any missing exponents:  P(x) = -2x4 + 4x3 + 0x2 -x + 6

 

Write all the coefficients in a line (including the constant) with the number being solved for off to the left:

 

Bring down the first coefficient (-2), multiply it by the term in question (2), carry the product up under

the 2nd coefficient and then add down (4-4=0), carry up the sum and repeat process across.  The last

sum is the answer for P(2)

 

 

(2)   -2    4    0    -1    6

            -4    0      0  -2

      __________________

       -2   0    0     -1   4

 

P(2) = 4

 

check the answer:  P(2) = -2(24) + 4(23) -2 + 6 = -2(16) +4(8) + 4 = 4    Our answer is correct

 

The quotient is what we would bet by dividing the original equation by the polynomial (x-2). The

answer is given by the bottom numbers which will begin an one lower exponent than the original.

Quotient is:   -2x3 + 0x2 + 0x -1 = 2x3 - 1

The remainder is:   4/(x-2)

6 0
3 years ago
Simplify the monomials
Zielflug [23.3K]

Answer:

(4a)^{-3}*^{-4}

x^m*x^n=x^{m+n}

(4a)^{-3}*a^{-4}

(-4)^{-3}\: a^{-3}*a^{-4}

4^{-3}\:a^{-3-4}

\cfrac{1}{4^3} \:a^{-7}

\boxed{\frac{1}{64a^{7}} }

~

(3xy)^2*(-4x^3y^2)^3

9x^2y^2*(64x^9y^6)

\boxed{-576x^{11}y^8}

~

(4a^{-1}b^5c^{-3})^3

(4)^3(a^{-1})^3(b^5)^3(c^{-3})^3

(4*4*4)\:a^{-1*3}\:b^{5*3}\:c^{-3*3}

64\:a^{-3}b^{15}c^{-9}

\boxed{\frac{64b^{15}}{a^3c^9}}

7 0
2 years ago
I’m a bit confused in learning about congruent triangles
Tems11 [23]
Okay and... what are you confused about
8 0
2 years ago
Equation of the line that passes through the point (4,–5) and is perpendicular to the line y=1/3x+5
fomenos

Answer:

y=1/3x+5 hope it helps you

5 0
2 years ago
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