Triangle PQR is transformed to triangle P'Q'R'. Triangle PQR has vertices P(3, −6), Q(0, 9), and R(−3, 0). Triangle P'Q'R' has vertices P'(1, −2), Q'(0, 3), and R'(−1, 0). Plot triangles PQR and P'Q'R' on your own coordinate grid. Part A: What is the scale factor of the dilation that transforms triangle PQR to triangle P'Q'R'? Explain your answer. (4 points) Part B: Write the coordinates of triangle P"Q"R" obtained after P'Q'R' is reflected about the y-axis. (4 points) Part C: Are the two triangles PQR and P''Q''R'' congruent? Explain your answer. (2 points)
@BestBoiKirby
i am here
one q before i start
is this is a test question?
It is for a review of a test. But I cant figure this stupid question out and it doesnt even give me the grid to try and solve it
ok, so do you have grid paper? or do you have enough art skills to make a decent looking grid?
I could try but I dont know the coordinates or anything
well, heres what you should do, ill list it srep by step
"srep", lol
make a giant plus sine for the base of ur grid
sign*
I know that part equals 2 but part B and C make no sense
huh? you just completely threw me off
part A equals 2
I got that far but as for part B and C. It doesnt make any sense
if im being 1000% honest, i feel like you should ping someone better at math than me, i can try, but i feel you would get the best results out of someone like mira angel
do you want me to ping her or are you ok with meh bad explanations
You can ping her if you want
shes offline TwT
@darkknight, you're good at math, can you help him and give a good explanation?
or @silvernight269
bring all the knights
Get some graph paper out or use Desmos to graph the triangles
Why is this hard?! Like why cant math just be easy
\[D = \sqrt (x _{2-} x _{1})^2 + (y _{2} - y _{1})^2\] You can use this formula to calculate the distance between points if you can't tell on the graph
?????????
TBH Iḿ terrible at math. I understand some of it but the rest is just way to hard. I dont understand the formulas and all that
Okay give me a min
well, offline, lol
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