Answer:
Given : PS = QR
∠PRS = ∠RPQ
PR = PR (On the same base)
Proof / Reason :
∠PRS = ∠RPQ (R)
PS = QR (H)
PR = PR (S)
Therefore, Triangle PRS is congruent to Triangle RPQ
∠PSR = ∠RQP By C.P.C.T. ( Congruent Parts Of Congruent Triangles)
Thanks!
Rise over run so it would be 5 over -2.5 ?? I could be wrong:((
Answer:
m∠FXD = 36°
EG = 10
m∠FZG = 24°
Step-by-step explanation:
In triangle XYZ, G is the incenter of the triangle.
Since, m∠FXG = 18°
And m∠FXD = 2(m∠FXG)
= 2 × 18°
= 36°
Since, point G is equidistant from all sides (Property of incenter of a triangle)
Therefore, DG = EG = GF = 10
Since, m∠X + m∠Y + m∠Z = 180°
2(m∠FXG) + m∠DYE + 2(m∠FZG) = 180°
2(18)° + 96° + 2(m∠FZG) = 180°
2(m∠FZG) = 180° - 132°
m∠FZG = 24°
Answer:
x=2 and y= -3
Step-by-step explanation:
This is a simultaneous equation. To solve this type of equation, there are three methods; substitution, Elimination and Graphical.
But here, we would be using the substitution method.
3x-y=9 Equation 1
2x-y=7. Equation 2
Getting y from equation 2, we have
-y= 7-2x
Multiply both sides by -
y= 2x-7 Equation 3
Substituting y for 2x-7 in equation 1, we have
3x- (2x-7)=9
3x-2x+7=9
x+7=9
x=9-7
x=2
Substituting x as 2 in equation 3
y=2x-7
y= 2(2)-7
y= 4-7
y= -3
Answer:
16272 is the answer.......
