Transversal t cuts parallel lines r and s. which angles must be congruent to < 2?

∠2 ≅ ∠6 ≅ ∠7 ≅ ∠3

Step-by-step explanation:

Lines r and s are the parallel lines and t then we have to find the angles congruent to ∠2.

∠2 ≅ ∠3

[Since ∠2 and ∠3 are vertically opposite angles.

∠2 ≅ ∠6

[Since ∠2 and ∠6 are corresponding angles]

∠2 ≅ ∠7

Since ∠2 ≅ ∠6

and ∠6 ≅ ∠7 [vertically opposite angles]

So ∠2 ≅ ∠7  [Transitive property of congruence]

Therefore, ∠2 ≅ ∠6 ≅ ∠7 ≅ ∠3 will be the answer.

456567546657689765432567890-9876543567890-987657890

step-by-step explanation:

the quotient is 4x^2 - 27x + 167 with a remainder of -996

step-by-step explanation:

in synthetic division, use -6 as the divisor and use the coefficients {4, -3, 5, 6}:

-6     )   4     -3     5     6

                -24   162 -1002

          4   -27   167   -996

thus, the quotient is 4x^2 - 27x + 167 with a remainder of -996.

note:   your "4x°-3x+5x+6" is ambiguous / incomplete.   i had to assume that you meant '4x^3 - 3x^2 + 5x + 6."