Bill Jorden has a 48-month installment loan with a fixed monthly payment of

There are 12 different ways she could take a penny, a nickle, and a dime out of here pocket and only pick 2. What i did was i took 2 and 3 and multiplied them. because i can pick 2 coins, and there are 3 coins in total. After i multiplied them i got 6, then i got 2 and multiplied it by 6 and came out with 12. The reason why i did that is because there are 6 outcomes so far then you have to multiply them by 2 to get all of your complete outcomes. I hope i could help!!! Please give me brainliest!!! Have a great day =D!!!

3 0

4 years ago

Instructions: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

Akimi4 [234]

to compare the triangles, first we will determine the distances of each side

Distance = ((x2-x1)^2+(y2-y1)^2)^0.5
Solving

∆ABC A(11, 6), B(5, 6), and C(5, 17)

AB = 6 units BC = 11 units AC = 12.53 units
∆XYZ X(-10, 5), Y(-12, -2), and Z(-4, 15)
XY = 7.14 units YZ = 18.79 units XZ = 11.66 units

∆MNO M(-9, -4), N(-3, -4), and O(-3, -15).

MN = 6 units NO = 11 units MO = 12.53 units
∆JKL J(17, -2), K(12, -2), and L(12, 7).
JK = 5 units KL = 9 units JL = 10.30 units
∆PQR P(12, 3), Q(12, -2), and R(3, -2)
PQ = 5 units QR = 9 units PR = 10.30 units
Therefore
we have the ∆ABC and the ∆MNO 
with all three sides equal ---------> are congruent
we have the ∆JKL and the ∆PQR
with all three sides equal ---------> are congruent

let's check

Two plane figures are congruent if and only if one can be obtained from the other by a sequence of rigid motions (that is, by a sequence of reflections, translations, and/or rotations).

1) If ∆MNO ---- by a sequence of reflections and translation --- It can be obtained ------->∆ABC

then ∆MNO ≅ ∆ABC

a) Reflexion (x axis)

The coordinate notation for the Reflexion is (x,y)---- >(x,-y)

∆MNO M(-9, -4), N(-3, -4), and O(-3, -15).

M(-9, -4)-----------------> M1(-9,4)

N(-3, -4)------------------ > N1(-3,4)

O(-3,-15)----------------- > O1(-3,15)

b) Reflexion (y axis)

The coordinate notation for the Reflexion is (x,y)---- >(-x,y)

∆M1N1O1 M1(-9, 4), N1(-3, 4), and O1(-3, 15).

M1(-9, -4)-----------------> M2(9,4)

N1(-3, -4)------------------ > N2(3,4)

O1(-3,-15)----------------- > O2(3,15)

c) Translation

The coordinate notation for the Translation is (x,y)---- >(x+2,y+2)

∆M2N2O2 M2(9,4), N2(3,4), and O2(3, 15).

M2(9, 4)-----------------> M3(11,6)=A

N2(3,4)------------------ > N3(5,6)=B

O2(3,15)----------------- > O3(5,17)=C

∆ABC A(11, 6), B(5, 6), and C(5, 17)

∆MNO reflection------- > ∆M1N1O1 reflection---- > ∆M2N2O2 translation -- --> ∆M3N3O3

The ∆M3N3O3=∆ABC

Therefore ∆MNO ≅ ∆ABC - > check list

2) If ∆JKL -- by a sequence of rotation and translation--- It can be obtained ----->∆PQR

then ∆JKL ≅ ∆PQR

d) Rotation 90 degree anticlockwise

The coordinate notation for the Rotation is (x,y)---- >(-y, x)

∆JKL J(17, -2), K(12, -2), and L(12, 7).

J(17, -2)-----------------> J1(2,17)

K(12, -2)------------------ > K1(2,12)

L(12,7)----------------- > L1(-7,12)

e) translation

The coordinate notation for the translation is (x,y)---- >(x+10,y-14)

∆J1K1L1 J1(2, 17), K1(2, 12), and L1(-7, 12).

J1(2, 17)-----------------> J2(12,3)=P

K1(2, 12)------------------ > K2(12,-2)=Q

L1(-7, 12)----------------- > L2(3,-2)=R

∆PQR P(12, 3), Q(12, -2), and R(3, -2)

∆JKL rotation------- > ∆J1K1L1 translation -- --> ∆J2K2L2=∆PQR

Therefore ∆JKL ≅ ∆PQR - > check list

6 0

3 years ago