2 years ago

Can anyone help me with this e?I need the answer quickly.(FINALS ON MONDAY)

Mathematics

2 answers:

Furkat [3]2 years ago

6 0

x^2+ 1 /x^2 +2−3x− 3/ x

x^5−3x^4+2x^3−3x^2+x /x^3

x^4−3x^3+2x^2−3x+1 / x2

Nesterboy [21]2 years ago

4 0

here is a photo for you

mark me as brainliest if this helped

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Step-by-step explanation:

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2 years ago

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2 years ago

Read 2 more answers

What is the expression in radical form?<br><br> (4x3y2)310

sertanlavr [38]

Given:

Consider the given expression is

$(4x^3y^2)^{\frac{3}{10}}$

To find:

The radical form of given expression.

Solution:

We have,

$(4x^3y^2)^{\frac{3}{10}}=(2^2)^{\frac{3}{10}}(x^3)^{\frac{3}{10}}(y^2)^{\frac{3}{10}}$

$(4x^3y^2)^{\frac{3}{10}}=(2)^{\frac{6}{10}}(x)^{\frac{9}{10}}(y)^{\frac{6}{10}}$

$(4x^3y^2)^{\frac{3}{10}}=(2)^{\frac{3}{5}}(x)^{\frac{9}{10}}(y)^{\frac{3}{5}}$

$(4x^3y^2)^{\frac{3}{10}}=\sqrt[5]{2^3}\sqrt[10]{x^9}\sqrt[5]{y^3}$ $[\because x^{\frac{1}{n}}=\sqrt[n]{x}]$

$(4x^3y^2)^{\frac{3}{10}}=\sqrt[5]{8y^3}\sqrt[10]{x^9}$ $[\because x^{\frac{1}{n}}=\sqrt[n]{x}]$

Therefore, the required radical form is $\sqrt[5]{8y^3}\sqrt[10]{x^9}$ .

8 0

3 years ago

Can anyone help me with this e?I need the answer quickly.(FINALS ON MONDAY)​

Can anyone help me with this e?I need the answer quickly.(FINALS ON MONDAY)