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3 years ago

I am lost on "Graph y< -3|x-1| please help!

Mathematics

1 answer:

Sindrei [870]3 years ago

6 0

1) |x| is the absolute value function. Its vertex is at (0,0).

2) |x-1| has the same graph as does y = |x|, except that the vertex is shifted 1 unit to the left.

3) -3|x-1| has the same graph as the previous result, except that we must reflect the previous graph in the x-axis.

4) Replace the solid lines you used in this graph with dashed ones.

5) Shade the area beneath y = -3|x-1|

In summary:

the desired graph is the shaded area below the inverted v-shaped graph of

y = -3|x-1|. Its vertex is at (-1,0).

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Line segment ST is dilated to create line segment S'T' using the dilation rule DQ,2. 25. Point Q is the center of dilation. Line

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The distance between points S' and S is x= 2.5 units.

<h2>Given that</h2>

Line segment ST is dilated to create line segment S'T' using the dilation rule DQ 2. 25.

Point Q is the center of dilation.

Line segment ST is dilated to create line segment S prime T prime.

The length of QT is 1. 2 and the length of QS is 2.

The length of SS prime is x and the length of TT prime is 1. 5.

<h3>We have to determine</h3>

What is x, the distance between points S' and S?

<h3>According to the question</h3>

Line segment ST is dilated to create line segment S'T' using the dilation rule DQ, 2.25.

Also, SQ = 2 units, TQ = 1.2 units, TT'=1.5, SS' = x.

Since the line ST is dilated to S'T' with the center of dilation Q, the triangles STQ and S'T'Q must be similar.

We know that the corresponding sides of two similar triangles are proportional.

So, from ΔSTQ and ΔS'T'Q.

$\dfrac{SQ}{S'Q} = \dfrac{TQ}{T'Q}\\
\\
\dfrac{SQ}{SQ+SS} = \dfrac{TQ}{TQ+TT'}\\
\\ 
\dfrac{2}{2+x} = \dfrac{1.2}{1.2+1.5}\\\rm 
\\
\dfrac{2}{2+x} = \dfrac{1.2}{2.7}\\\\ \dfrac{2}{2+x} = \dfrac{12}{27}\\\\2(27) = (2+x) 12\\\\ 54 = 24 + 12x\\
\\
12x = 54-24\\
\\
12x=30\\
\\
x = \dfrac{30}{12}\\
\\
x = 2.5$

Hence, the distance between points S' and S is x= 2.5 units.

To know more about Pythagoras Theorem click the link given below.

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