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2 years ago

HELP ASAP!!! What system of inequalities is shown in the graph

Mathematics

1 answer:

storchak [24]2 years ago

4 0

They need to be number C plus 3-2:7

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A. Clara drew the figure correctly and correctly identified the four-sided figure as a quadrilateral.

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The figure drawn is attached.

This is a quadrilateral.

Clara did it correctly and named properly.

Correct option is A.

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Which equation could have been used to create this function table? A. y = x + 1 B. y = x + 6 C. y = 5x D. y = 3x

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AlladinOne [14]

Answer:

$\left(2\:\sqrt{3\:+\:3\:\sqrt{2}\:\:}\:\right)\:^2=4(3+3\sqrt{2})$

Step-by-step explanation:

Considering the radical expression

$\left(2\:\sqrt{3\:+\:3\:\sqrt{2}\:\:}\:\right)\:^2$

$\mathrm{Apply\:exponent\:rule}:\quad \left(a\cdot \:b\right)^n=a^nb^n$

$=2^2\left(\sqrt{3+3\sqrt{2}}\right)^2.....[A]$

Simplifying

$\left(\sqrt{3+3\sqrt{2}}\right)^2$

$\mathrm{Apply\:radical\:rule}:\quad \sqrt{a}=a^{\frac{1}{2}}$

$=\left(\left(3+3\sqrt{2}\right)^{\frac{1}{2}}\right)^2$

$\mathrm{Apply\:exponent\:rule}:\quad \left(a^b\right)^c=a^{bc}$

$=\left(3+3\sqrt{2}\right)^{\frac{1}{2}\cdot \:2}$

$=3+3\sqrt{2}$ As $\frac{1}{2}\cdot \:2=1$

So, putting $3+3\sqrt{2}$ into Equation [A]

$=2^2\left(\sqrt{3+3\sqrt{2}}\right)^2.....[A]$

$\left(\sqrt{3+3\sqrt{2}}\right)^2=3+3\sqrt{2}$

So, Equation [A] becomes

$=2^{2}(3+3\sqrt{2})$

$=4(3+3\sqrt{2})$

Therefore,

$\left(2\:\sqrt{3\:+\:3\:\sqrt{2}\:\:}\:\right)\:^2=4(3+3\sqrt{2})$

Keywords: radical expression

Learn more radical expression from brainly.com/question/11624221

#learnwithBrainly

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3 years ago

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