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3 years ago

1.) 6x + 3x^2 - x - x^2

Mathematics

2 answers:

svlad2 [7]3 years ago

7 0

Answer:

the answer is x=1/3y+1 y=3x−3

but wat the question?

shutvik [7]3 years ago

5 0

$\large \mathfrak{Solution : }$

1. 6x + 3x² - x - x²

3x² - x² + 6x - x

2x² + 5x

2. 5 + 2x + y + 2x - 1

2x + 2x + y + 5 - 1

4x + y + 4

3. 2x + 2y + x² - x + x²

x² + x² + 2x - x + 2y

2x² + x + 2y

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Please solve this, will rate 5 stars and mark as STAR!

Nina [5.8K]

Answer:

$\boxed{5 \cdot \sqrt{2} \cdot \sqrt[6]{5} }$

Step-by-step explanation:

$\sqrt[3]{250} \cdot \sqrt{\sqrt[3]{10} }$

$\sqrt{\sqrt[3]{10} } \implies (10^\frac{1}{3} )^\frac{1}{2} =10^\frac{1}{6} =\sqrt[6]{10}$

$\therefore \sqrt{\sqrt[3]{10} }=\sqrt[6]{10}$

$\text{Solving }\sqrt[3]{250} \cdot \sqrt{\sqrt[3]{10} }$

$250=2 \cdot 5^3$

$\sqrt[3]{250}=\sqrt[3]{2\cdot 5^3}=5 \sqrt[3]{2}$

Once

$\sqrt[6]{2} \cdot \sqrt[6]{5} = \sqrt[6]{10}$

We have

$5 \sqrt[3]{2} \cdot \sqrt[6]{2} \cdot \sqrt[6]{5}$

We can proceed considering the common base of exponentials

$\sqrt[3]{2} \cdot \sqrt[6]{2} = 2^{\frac{1}{3}} \cdot 2^{\frac{1}{6} } = 2^{\frac{3}{6} } = 2^{\frac{1}{2} }=\sqrt{2}$

Therefore,

$5 \sqrt[3]{2} \cdot \sqrt[6]{2} \cdot \sqrt[6]{5} = 5 \cdot \sqrt{2} \cdot \sqrt[6]{5}$

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3 years ago

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Step-by-step explanation:

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Answer:

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