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3 years ago

Which point is the vertex for the graph of y = |x| + 2?

Mathematics

2 answers:

Tasya [4]3 years ago

4 0

The vertex is when x=0, so (0, 2) is the answer.

Marysya12 [62]3 years ago

3 0

Hey, it's (0,2) because the absolute value of zero, as x, is still zero and then + 2 = 2, for y

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3 years ago

On a coordinate plane, triangle A B C is shown. Point A is at (0, 0), point B is at (3, 4), and point C is at (3, 2). What is th

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The area of triangle for the given coordinates is 1.5 $\sqrt{4.6}$

Step-by-step explanation:

Given coordinates of triangles as

A = (0,0)

B = (3,4)

C = (3,2)

So, The measure of length AB = a = $\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}$

Or, a = $\sqrt{(3-0)^{2}+(4-0)^{2}}$

Or, a = $\sqrt{9+16}$

Or, a = $\sqrt{25}$

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The measure of length BC = b = $\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}$

Or, b = $\sqrt{(3-3)^{2}+(2-4)^{2}}$

Or, a = $\sqrt{0+4}$

Or, b = $\sqrt{4}$

∴ b = 2 unit

And

So, The measure of length CA = c = $\sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}$

Or, c = $\sqrt{(3-0)^{2}+(2-0)^{2}}$

Or, c = $\sqrt{9+4}$

Or, c = $\sqrt{13}$

∴ c = $\sqrt{13}$ unit

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A = $\sqrt{s\times (s-a)\times (s-b)\times (s-c)}$

and s = $\frac{a+b+c}{2}$

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Or. s = $\frac{7+\sqrt{13}}{2}$

So, A = $\sqrt{(\frac{(7+\sqrt{13})}{2})\times ((\frac{(7+\sqrt{13})}{2})-5)\times (\frac{7+\sqrt{13}}{2}-2)\times (\frac{7+\sqrt{13}}{2}-\sqrt{13})}$

Or, A = $\sqrt{(\frac{(7+\sqrt{13})}{2})\times (\frac{(\sqrt{13}-3)}{2})\times (\frac{4+\sqrt{13}}{2})\times (\frac{7-\sqrt{13}}{2})}$

Or, A = $\frac{3}{2}$ × $\sqrt{1+\sqrt{13} }$

∴ Area of triangle = 1.5 $\sqrt{4.6}$

Hence The area of triangle for the given coordinates is 1.5 $\sqrt{4.6}$ Answer

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