Select the correct answer from each drop-down menu. ∆ABC has vertices at A(11, 6), B(5, 6), and C(5, 17).
∆XYZ has vertices at X(-10, 5), Y(-12, -2), and Z(-4, 15).
∆MNO has vertices at M(-9, -4), N(-3, -4), and O(-3, -15).
∆JKL has vertices at J(17, -2), K(12, -2), and L(12, 7).
∆PQR has vertices at P(12, 3), Q(12, -2), and R(3, -2).

can be shown to be congruent by a sequence of reflections and translations.
can be shown to be congruent by a single rotation.

What kind of math is this wish I could help

i dont now

Step-by-step explanation:

answer: 5

step-by-step explanation:

answer: solve 5\cdot 2^x=2405⋅2  

x

=2405, dot, 2, start superscript, x, end superscript, equals, 240.

solution

to solve for xxx, we must first isolate the exponential part. to do this, divide both sides by 555 as shown below. we do not multiply the 555 and the 222 as this goes against the order of operations!

\begin{aligned} 5\cdot 2^x& =240 2^x& =48 \\ \end{aligned}  

5⋅2  

x

2  

x

​    

=240

=48

​

step-by-step explanation: