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zvonat [6]
2 years ago
15

What is the radical form of the expression (2x^4y^5)^3/8

Mathematics
1 answer:
s344n2d4d5 [400]2 years ago
8 0

We are given expression: (2x^4y^5)^{3/8}

Let us distribute 3/8 over exponents in parenthesis, we get

(2^{3/8}x^{4\times 3/8}y^{5\times 3/8}) = (2^{3/8}x^{12/8}y^{15/8})

= (2^{3/8}x^{1\frac{4}{8}} y^{1\frac{7}{8}} )

We can get x and y out of the radical because, we get whlole number 1 for x and y exponents for the mixed fractions.

So, we could write it as.

(2^{3/8}x^{1\frac{4}{8}} y^{1\frac{7}{8}} ) = xy(2^{\frac{3}{8} }x^{\frac{4}{8}} y^{\frac{7}{8}} )

Now, we could write inside expression of parenthesis in radical form.

xy\sqrt[8]{2x^{3}x^4y^7}

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Round off 3409725 to nearest 10​
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Answer:

10000

Step-by-step explanation:

to know if this answer is correct you have to know the rounding rules:

If the number you are rounding is followed by 5, 6, 7, 8, or 9, round the number up. Example: 38 rounded to the nearest ten is 40.

If the number you are rounding is followed by 0, 1, 2, 3, or 4, round the number down. Example: 33 rounded to the nearest ten is 30.

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1 year ago
Mapiya writes a series of novels. She earned \$75{,}000$75,000dollar sign, 75, comma, 000 for the first book, and her cumulative
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Answer:

E(n)=75000 \times 2^n

Complete question:

write a function that gives mapiyas cumulative earnings E(n), in dollars when she has written n sequel's

Step-by-step explanation:

According to the question, she earned $75000 for the first book.

Also,We are  given that her cumulative earnings double with each sequel that she writes.

Assuming she has written n sequel's

Now since we are given that her cumulative earnings double with each sequel

So, her initial earning will be 2^n times

So, her earning will be : 75000 \times 2^n

Now we are given that cumulative earnings is denoted by E(n)

So, the function becomes :E(n)=75000 \times 2^n

Hence a function that gives Mapiya's cumulative earnings E(n), in dollars when she has written n sequel's  is  E(n)=75000 \times 2^n

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if AnC = AnC does not imply that B = C.

if A = {1,2}, B = {1,2,3,4,5} and C = {1,2,3}

then AnC = {1,2} and AnB = {1,2} but B and C are different.

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