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DerKrebs [107]
3 years ago
13

What is the simplified form of the following expression? 7 (RootIndex 3 StartRoot 2 x EndRoot) minus 3 (RootIndex 3 StartRoot 16

x EndRoot) minus 3 (RootIndex 3 StartRoot 8 x EndRoot) Negative 5 (RootIndex 3 StartRoot 2 x EndRoot) 5 (RootIndex 3 StartRoot x EndRoot) RootIndex 3 StartRoot 2 x EndRoot minus 6 (RootIndex 3 StartRoot x EndRoot) Negative (RootIndex 3 StartRoot 2 x EndRoot) minus 6 (RootIndex 3 StartRoot x EndRoot)
Mathematics
2 answers:
stiv31 [10]3 years ago
5 0

Answer:

g(x) = RootIndex 3 StartRoot x + 2 EndRoot

Step-by-step explanation:

Answer for e2020

B

11111nata11111 [884]3 years ago
3 0

Answer:

c

Step-by-step explanation:

dont use edge, and thats my only explanation

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4/15 of the sand. Multiply 4/5 by 1/3.
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I The cost CC (in dollars) of making n watches is represented by C=15n+85C=15n+85. How many watches are made when the cost is $3
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C = 15n + 85....when C = 385

385 = 15n + 85
385 - 85 = 15n
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300/15 = n
20 = n <== 20 watches were made
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What is the solution to the equation x + 4.7 = 6.3?
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Viefleur [7K]

Answer:

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5 0
3 years ago
Suppose that a computer software company has 30
LekaFEV [45]

Answer: \dfrac{30!}{6!(24)!}

Step-by-step explanation:

Given : The total number of programmers in the company = 30

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Since the order of selecting them does not matters , therefore we use combinations.

The number of combinations of r things taken from n things is given by :-

^nC_r=\dfrac{n!}{r!(n-r)!}

here, n= 30 and r= 6

So the number of different ways to form they could select a group of 6 would be ^{30}C_{6}=\dfrac{30!}{6!(30-6)!}

=\dfrac{30!}{6!(24)!}\\\\=\dfrac{30\times29\times28\times27\times26\times25\times24!}{(720)24!}=593775

i.e. Total ways =593775

In terms of factorials , the number of total ways to form they could select a group of 6 is \dfrac{30!}{6!(24)!} .

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