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3 years ago

Plz help me get the answer ASAP

Mathematics

2 answers:

Anvisha [2.4K]3 years ago

8 0

Answer:

Step-by-step explanation:

$\dfrac{1}{2}c>-1\qquad\text{multiply both sides by 2}\\\\2\!\!\!\!\diagup^1\cdot\dfrac{1}{2\!\!\!\!\diagup_1}c>(2)(-1)\\\\c>-2$

Georgia [21]3 years ago

7 0

Answer: c > -2

<u>Explanation:</u>

$\frac{1}{2}c > -1$

(2) $\frac{1}{2}c > -1(2)$

c > -2

Graph: -2 -----------→

Interval Notation: [-2, ∞)

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3 years ago

Consider the table below.

Licemer1 [7]

Answer:

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Step-by-step explanation:

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3 years ago

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2 years ago

Read 2 more answers

Can I please get some help with this

Marizza181 [45]

After solving $\sqrt[5]{128x^8y^2}$ we get $2x\sqrt[5]{4}\sqrt[5]{x^3}\sqrt[5]{y^2}$

Step-by-step explanation:

We need to solve $\sqrt[5]{128x^8y^2}$

Applying the rule: $\sqrt[a]{xy}=\sqrt[a]{x}.\sqrt[a]{y}$

$\sqrt[5]{128x^8y^2}\\=\sqrt[5]{128}\sqrt[5]{x^8}\sqrt[5]{y^2}\\We\,\,know\,\,that\,\,128=2\times2\times2\times2\times2\times2\times2=2^7\,\,or\,\,2^5.2^2\\=\sqrt[5]{2^5.2^2}\sqrt[5]{x^5.x^3}\sqrt[5]{y^2}\\=(2^5)^{\frac{1}{5}}\sqrt[5]{2^2}\sqrt[5]{x^5}\sqrt[5]{x^3}\sqrt[5]{y^2}\\=2x\sqrt[5]{4}\sqrt[5]{x^3}\sqrt[5]{y^2}$

So, After solving $\sqrt[5]{128x^8y^2}$ we get $2x\sqrt[5]{4}\sqrt[5]{x^3}\sqrt[5]{y^2}$

Keywords: Solving with Exponents

Learn more about Solving with Exponents at:

#learnwithBrainly

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2 years ago