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Wittaler [7]
3 years ago
13

Write each expression in a simpler form that is equivalent to the given expression. Let F be a nonzero number. f-4

Mathematics
2 answers:
guajiro [1.7K]3 years ago
8 0

Answer:

\large \boxed{f-4}

Step-by-step explanation:

f-4

\sf f \ is \ a \ nonzero \ number.

\sf The \ expression \ cannot \ be \ simplified \ further.

natima [27]3 years ago
7 0

Answer:

f-4

Step-by-step explanation:

f-4  cannot be simplified

This is the simplest form

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it has to be irrational

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Write a linear function?
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y= 1/2x -5/2

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Kamran Company produces various types of fertilizer. No beginning units in process or finished units were on hand on January 1,
Sergeeva-Olga [200]

Answer:

Cost of Goods Manufactured   900,000

Step-by-step explanation:

Kamran Company

Cost of Goods Manufactured.

Total Materials Used  Rs 300,000

Direct Materials = 75% of Rs 300,000=  Rs 225,000

Direct Labor= 60% of 350,000=  Rs 210,000

Prime Cost = 435,000

Factory Overhead

Indirect Materials = 75,000

Indirect Labor = 90,000

Heat, light and power Rs. 115,000

Depreciation 78,000

Factory Taxes 65,000

Repairs and Maintenance 42,000

Cost of Good Available for Manufacture : 900,000

Add opening WIP  --- Nil

Less Closing WIP ---Nil

Cost of Goods Manufactured   900,000

Selling Expenses were Rs. 80,000 general and administrative expenses were Rs. 50,000 are used in making Profit & Loss or the income statement.

Beginning and Ending Finished Goods are used in the Cost of Goods Sold Statement.

4 0
3 years ago
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Answer:

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3 years ago
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In the expansion of (1/ax +2ax^2)^5 the coefficient of x is five. Find the value of the constant a.
DedPeter [7]

Answer:

80x⁴

Step-by-step explanation:

(\frac{1}{ax} + 2ax^2)^5 = 5C_0(\frac{1}{ax})^5(2ax^2)^0 + 5C_1(\frac{1}{ax})^4(2ax^2)^1 + 5C_2(\frac{1}{ax})^3 (2ax^2)^2

                           + 5C_3 (\frac{1}{ax})^2(2ax^2)^3 + 5C_4(\frac{1}{ax})^1(2ax^2)^4 + 5C_5(\frac{1}{ax})^0(2ax^2)^5

5C_0(\frac{1}{ax})^5(2ax^2)^0  =1 \times (\frac{1}{ax})^5 \times 1 = \frac{1}{a^5x^5}\\\\5C_1(\frac{1}{ax})^4(2ax^2)^1  = 5 \times (\frac{1}{ax})^4 \times (2ax^2)^1 = 10 ax^2 \times \frac{1}{a^4x^4} = \frac{10}{a^3x^2}\\\\5C_2 (\frac{1}{ax})^3 (2ax^2)^2= 10 \times (\frac{1}{ax})^3 \times (2ax^2)^2 = 10 \times \frac{1}{a^3x^3} \times 4a^2x^4 = \frac{40x}{a}\\\\5C_3 (\frac{1}{ax})^2 (2ax^2)^3 = 10 \times (\frac{1}{ax})^2 \times (2ax^2)^3 = 10 \times \frac{1}{a^2x^2} \times 8a^3 x^6 = 80ax^4\\\\

5C_4(\frac{1}{ax})^1(2ax^2)^4 = 5 \times \frac{1}{ax} \times 16a^4x^8 = 80a^3x^7\\\\5C_5(\frac{1}{ax})^0(2ax^2)^5 = 1 \times 1 \times 32a^5x^{10}

The fourth term of the expansion has the constant a,

the coefficient of a is 80x⁴

6 0
3 years ago
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