615 - 6 = 609
963 - 6 = 957
609=3×3×29
957=3×11×29
HCF=3×29=87
The largest number that divides 615 and 963 leaving remainder 6 in both numbers is 87.
Angles ∠MOK and ∠NOL are identical by opposite angle theorem.
∠M = ∠N through given information
∠MKO and ∠NLO are congruent. Since the other two angles of each triangle are equivalent, the last angle must be equivalent to the last angle of the other triangle. (180 - ∠1 -∠2 = ∠3); you could also prove ∠MKO = ∠NLO by stating that the opposite sides of the two angles are congruent.
ΔMKO and ΔNLO by Angle-Angle-Side theorem
since ΔMKO and ΔNLO, all corresponding parts of both triangles are congruent, so therefore MO ≅ NO
The answer is B. It does not represent a function because a vertical line will cross the graph at more than one point
They will have the same solution because the first equation of system b
is obtained by adding the first equation of system a to 4 times the
second equation of system a.
