√131 is between 11 and 12? true or false

Mathematics

2 answers:

guajiro [1.7K]3 years ago

6 0

True
11 is √121
12 is √144
So √131 is in between 11 and 12

kirza4 [7]3 years ago

4 0

Answer:

True

Step-by-step explanation:

$11^2=121\\12^2=144\\121$

Hence, $\sqrt{131}$ is in between 11 and 12

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Adrianna is attempting to write an inequality from a graph. She knows that the dash boundary line passes through the points (27,

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we know the line passes through those points, so let's find the EQUATion of it firstly.

$\bf (\stackrel{x_1}{27}~,~\stackrel{y_1}{10})\qquad (\stackrel{x_2}{-45}~,~\stackrel{y_2}{-54}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{-54-10}{-45-27}\implies \cfrac{-64}{-72}\implies \cfrac{8}{9} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-10=\cfrac{8}{9}(x-27) \\\\\\ y-10=\cfrac{8}{9}x-24\implies y=\cfrac{8}{9}x-14$

so, that's its EQUATion, now, we know the (0,0) is a solution of this inequality, namely the (0,0) point lies in the "true region" or will be the "shaded region" of the inequality, let's plug those values,

$\bf y=\cfrac{8}{9}x-14\implies \stackrel{\textit{using x = 0, y = 0}}{0=\cfrac{8}{9}(0)-14}\implies 0=-14$