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.

answer: i forgot

step-by-step explanation:

answer: c

because if you look at the chart you can see a pattern and compare it to the question he simplify out your

but i think the formula is f(x)=6x

the number of possible outcomes that do not show a 1 on the first spinner and show the number 4 on the second spinner is 2.

spinner 1 spinner 2

1                     1

1                     2

1                     3

1                     4

2                   1

2                   2

2                   3

2                   4   = not 1, and 4

3                   1

3                   2

3                   3

3                   4 = not 1, and 4

only 2 outcomes have passed the given condition.

step-by-step explanation:

yes

step-by-step explanation:

so of course you want to start by plugging the 5 in which goes to 7(5)+3(5)=50 because 7*5=35 and 3*5=15 and then add them together, then plug in the 5 to the other equation as well 9(5)+5=50 because 9*5= 45 and then add 5 and so yes they are equivalent.