I really need help Dz,2 of X is (0-4) (2,-2) (6,2)

Mathematics

1 answer:

kvv77 [185]3 years ago

8 0

9514 1404 393

Answer:

(6,2)

Step-by-step explanation:

As is often the case with multiple-choice problems, you don't actually need to know the detailed working. You just need to know what the answer looks like.

When point X is dilated by a factor of 2 with point Z as the center of dilation, it will move to a location twice as far from Z. You can tell by looking at the graph that X' will be in the first quadrant, above and to the right of the location of X. The only sensible answer choice is ...

X' = (6, 2)

_____

<em>Additional comment</em>

X is a distance of X-Z = (4, 0) -(2, -2) = (2, 2) from Z Doubling that will put the image point a distance of 2(2, 2) = (4, 4) from Z. When this is added to Z, we find ...

X' = Z + (4, 4) = (2+4, -2+4) = (6, 2)

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What value of b will cause the system to have an infinite number of solutions?

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b must be equal to -6 for infinitely many solutions for system of equations $y = 6x + b$ and $-3 x+\frac{1}{2} y=-3$

<u>Solution: </u>

Need to calculate value of b so that given system of equations have an infinite number of solutions

$\begin{array}{l}{y=6 x+b} \\\\ {-3 x+\frac{1}{2} y=-3}\end{array}$

Let us bring the equations in same form for sake of simplicity in comparison

$\begin{array}{l}{y=6 x+b} \\\\ {\Rightarrow-6 x+y-b=0 \Rightarrow (1)} \\\\ {\Rightarrow-3 x+\frac{1}{2} y=-3} \\\\ {\Rightarrow -6 x+y=-6} \\\\ {\Rightarrow -6 x+y+6=0 \Rightarrow(2)}\end{array}$

Now we have two equations

$\begin{array}{l}{-6 x+y-b=0\Rightarrow(1)} \\\\ {-6 x+y+6=0\Rightarrow(2)}\end{array}$

Let us first see what is requirement for system of equations have an infinite number of solutions

If $a_{1} x+b_{1} y+c_{1}=0$ and $a_{2} x+b_{2} y+c_{2}=0$ are two equation

$\Rightarrow \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$ then the given system of equation has no infinitely many solutions.

In our case,

$\begin{array}{l}{a_{1}=-6, \mathrm{b}_{1}=1 \text { and } c_{1}=-\mathrm{b}} \\\\ {a_{2}=-6, \mathrm{b}_{2}=1 \text { and } c_{2}=6} \\\\ {\frac{a_{1}}{a_{2}}=\frac{-6}{-6}=1} \\\\ {\frac{b_{1}}{b_{2}}=\frac{1}{1}=1} \\\\ {\frac{c_{1}}{c_{2}}=\frac{-b}{6}}\end{array}$