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3 years ago

Does the order of variables matter when combining like terms in algebra

1 answer:

4 0

Answer: No
The order of variables does not matter when combining like terms in algebra, because of the commutative property.

Explanation:
Consider the addition of two algebraic expressions.
f(x) = x³ - 5x² + 3x - 2
g(x) = 6x³ + 10x² - 7x + 9

Add f(x) and g(x).
f(x) + g(x) = x³ - 5x² + 3x - 2 + (6x³ + 10x² - 7x + 9)
= x³ + 6x³ - 5x² + 10x² + 3x - 7x - 2 + 9
= 7x³ + 5x² - 4x + 7

Now add g(x) and f(x) in order to change the order of variable addition.
g(x) + f(x) = 6x³ + 10x² - 7x + 9 + (x³ - 5x² + 3x - 2)
= 6x³ + x³ + 10x² - 5x² - 7x + 3x + 9 - 2
= 7x³ + 5x² - 4x + 7

The two additions yield the same result, although the order of variable was reversed during the addition.

In general, algebraic operations are commutative.

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3 years ago

Braun Company has one service department and two operating (production) departments. Maintenance Department costs are allocated

vekshin1

Answer:

$122,600

Explanation:

Maintenance department cost = $27,000

Assembly department cost = $106,400

Square feet occupied by Milling department = 12,000

Square feet occupied by Assembly department = 18,000

Total square feet occupied by two production departments = Square feet occupied by Milling department + Square feet occupied by Assembly department

= 12,000 + 18,000

= 30,000

Maintenance department cost allocated to Assembly department = Maintenance department cost * (Square feet occupied by Assembly department / Total square feet occupied by two production departments)

= 27,000 * (18,000/30,000)

= 27,000 * 0.6

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3 years ago

A new highway is to be constructed.

baherus [9]

Answer:

Desing B

Total Present worth $ 502.485,35‬

Annual worth: $ 49,722.003

Option 2:

Total Present worth $ 666.441,33‬

Annual worth: $ 53,845.798

Explanation:

<h2>Desing A</h2>

$85 x 5280 = 448,800

$4 x 3 x 5280 = 63,360

$8000 x 4 = 32,000

total cost: 544,160‬

Annual cost:

11,800 + 300 = 12,100

PV of the annual maintenance:

$C \times \frac{1-(1+r)^{-time} }{rate} = PV\\$

C 12,100.00

time 16

rate 0.06

$12100 \times \frac{1-(1+0.06)^{-16} }{0.06} = PV\\$

PV $122,281.3328

Present worth:

total cost to construct 544,160‬ + maintenance $122,281.33 = 666.441,33

Annual worth:

$PV \div \frac{1-(1+r)^{-time} }{rate} = C\\$

PV 544,160

time 16

rate 0.06

$544160 \div \frac{1-(1+0.06)^{-16} }{0.06} = C\\$

C $ 53,845.798

<h2>Desing B</h2>

a mile is equivalent to 5,280 foot

paviment $40 x 5,280 = $ 211,200

sood ditched 2 per foot x 5,280 foot per mile x $1.40 = $ 12,038.4

pipe culvert 2,200 x 2 = 4,400

Total value to construct: 227.638,4‬

PV of maintenance:

replacement 2,400 x 2 = 4,800 (in 8 years)

$\frac{Maturity}{(1 + rate)^{time} } = PV$

Maturity $4,800.0000

time 8.00

rate 0.06000

$\frac{4800}{(1 + 0.06)^{8} } = PV$

PV 3,011.5794

maintenance $2,900

culverts 2 x $ 230 = $ 460

ditch 1.45 x 5,280 x 2 = $ 15,312

Total yearly cost: 18.672‬

PV of this annuity over 16 years:

$C \times \frac{1-(1+r)^{-time} }{rate} = PV\\$

C 18,672.00

time 16

rate 0.06

$18672 \times \frac{1-(1+0.06)^{-16} }{0.06} = PV\\$

PV $188,697.2765

PV of the replacement bituminous concrete

$\frac{Maturity}{(1 + rate)^{time} } = PV$

Maturity $211,200.0000

time 16.00

rate 0.06000

$\frac{211200}{(1 + 0.06)^{16} } = PV$

PV 83,138.0951

Present worth:

Total value to construct: $ 227.638,4‬ +

yearly cost PB $ 188,697.28 +

concrete replacement $ 83,138.0951

culvert replacement: $ 3,011.58

Total Present worth 502.485,35‬

Annual worth:

$PV \div \frac{1-(1+r)^{-time} }{rate} = C\\$

$502485.35 \div \frac{1-(1+0.06)^{-16} }{0.06} = C\\$

C $ 49,722.003

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3 years ago