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Rudiy27
1 year ago
7

a staright line has equation 3y-3x=4. Write down the equation of another straight line parallel to it.​

Mathematics
1 answer:
Doss [256]1 year ago
5 0

Answer:

Step-by-step explanation:

1 Factor out the common term 33.

3(y-x)=4.

3(y−x)=4.

2 Divide both sides by 33.

y-x=\frac{4.}{3}

y−x=

3

4.

​

3 Subtract yy from both sides.

-x=\frac{4.}{3}-y

−x=

3

4.

​

−y

4 Multiply both sides by -1−1.

x=-\frac{4.}{3}+y

x=−

3

4.

​

+y

5 Regroup terms.

x=y-\frac{4.}{3}

x=y−

3

4.

​

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Nataliya [291]

Answer:

1) Option A) is correct

The given rational exponent expression is not simplified correctly as a radical expression is

x^{\frac{7}{4}}=\sqrt[7]{x^4}

2)Option A) is correct

That is (729x^3y^{-18})^{\frac{1}{6}}=\frac{3\sqrt{x}}{y^3}

Step-by-step explanation:

1) Given that x^{\frac{7}{4}}=(\sqrt{x^7})^\frac{1}{4}

x^{\frac{7}{4}}=\sqrt[4]{x^7} is the correct answer but in the given problem they gave the RHS as wrong.

Therefore the given rational exponent expression is not simplified correctly as a radical expression is

x^{\frac{7}{4}}=\sqrt[7]{x^4}

2)Given that the rational exponent expression is (729x^3y^{-18})^{\frac{1}{6}}

To find it as a radical expression:

(729x^3y^{-18})^{\frac{1}{6}}=(729)^{\frac{1}{6}}(x^3)^{\frac{1}{6}}((y^{-18})^{\frac{1}{6}})

=3(x^{\frac{3}{6}})(y^{\frac{-18}{6}})

=3(x^{\frac{1}{2}})(y^{-3})

=\frac{3\sqrt{x}}{y^3}

Therefore (729x^3y^{-18})^{\frac{1}{6}}=\frac{3\sqrt{x}}{y^3}

Therefore Option A) is correct

That is (729x^3y^{-18})^{\frac{1}{6}}=\frac{3\sqrt{x}}{y^3}

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Answer:

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Step-by-step explanation:

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x  ≥  4

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