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3 years ago

Write the linear inequality shown in the graph. The gray area represents the shaded region

Mathematics

2 answers:

Sunny_sXe [5.5K]3 years ago

7 0

I believe the answer is y ≤ -3x + 5.
I have not learned inequalities yet so if I am wrong, then :/

According to m.a.t.h.w.a.y:

stira [4]3 years ago

3 0

(1,2)(2,-1)
slope = (-1 - 2) / (2 - 1) = -3/1 = -3

y = mx + b
slope(m) = -3
(1,2)...x = 1 and y = 2
sub and find b
2 = -3(1) + b
2 = -3 + b
2 + 3 = b
5 = b

equation is : y = -3x + 5.......u have a solid line, so there is an equal sign in the problem...it is shaded below the line, so it is less then.

ur inequality is : y < = -3x + 5

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Rationalize the denominator and simplify.

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If you're using the app, try seeing this answer through your browser: brainly.com/question/3151018

——————————

$\mathsf{\dfrac{3\sqrt{6}+5\sqrt{2}}{4\sqrt{6}-3\sqrt{2}}}$

Multiply and divide by the conjugate of the denominator, which is $\mathsf{4\sqrt{6}+3\sqrt{2}:}$

$\mathsf{=\dfrac{\big(3\sqrt{6}+5\sqrt{2}\big)\cdot \big(4\sqrt{6}+3\sqrt{2}\big)}{\big(4\sqrt{6}-3\sqrt{2}\big)\cdot \big(4\sqrt{6}+3\sqrt{2}\big)}}$

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$\mathsf{=\dfrac{\big(3\sqrt{6}+5\sqrt{2}\big)\cdot 4\sqrt{6}+ \big(3\sqrt{6}+5\sqrt{2}\big)\cdot 3\sqrt{2}}{\big(4\sqrt{6}-3\sqrt{2}\big)\cdot 4\sqrt{6}+ \big(4\sqrt{6}-3\sqrt{2}\big)\cdot 3\sqrt{2}}}\\\\\\ \mathsf{=\dfrac{12\cdot \big(\sqrt{6}\big)^2+20\sqrt{2}\cdot \sqrt{6}+9\cdot \sqrt{6}\cdot \sqrt{2}+15\cdot \big(\sqrt{2}\big)^2}{16\cdot \big(\sqrt{6}\big)^2-12\cdot \sqrt{2}\cdot \sqrt{6}+ 12\cdot \sqrt{6}\cdot \sqrt{2}-9\cdot \big(\sqrt{2}\big)^2}}\\\\\\ \mathsf{=\dfrac{12\cdot 6+20\sqrt{2\cdot 6}+9\sqrt{6\cdot 2}+15\cdot 2}{16\cdot 6-12\sqrt{2\cdot 6}+ 12\sqrt{6\cdot 2}-9\cdot 2}}\\\\\\ \mathsf{=\dfrac{72+20\sqrt{12}+9\sqrt{12}+30}{96-12\sqrt{12}+12\sqrt{12}-18}}$

$\mathsf{=\dfrac{72+29\sqrt{12}+30}{96-18}}\\\\\\ \mathsf{=\dfrac{102+29\sqrt{2^2\cdot 3}}{78}}\\\\\\ \mathsf{=\dfrac{102+29\cdot \sqrt{2^2}\cdot \sqrt{3}}{78}}\\\\\\ \mathsf{=\dfrac{102+29\cdot 2\sqrt{3}}{78}}$

$\mathsf{=\dfrac{102+58\sqrt{3}}{78}}\\\\\\ \mathsf{=\dfrac{\,\diagup\!\!\!\! 2\cdot \big(51+29\sqrt{3}\big)}{\diagup\!\!\!\! 2\cdot 39}}$

$\mathsf{=\dfrac{51+29\sqrt{3}}{39}}\quad\longleftarrow\quad\textsf{this is the answer.}$

I hope this helps. =)

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