Answer:
This is proved by ASA congruent rule.
Step-by-step explanation:
Given KLMN is a parallelogram, and that the bisectors of ∠K and ∠L meet at A. we have to prove that A is equidistant from LM and KN i.e we have to prove that AP=AQ
we know that the diagonals of parallelogram bisect each other therefore the the bisectors of ∠K and ∠L must be the diagonals.
In ΔAPN and ΔAQL
∠PNA=∠ALQ (∵alternate angles)
AN=AL (∵diagonals of parallelogram bisect each other)
∠PAN=∠LAQ (∵vertically opposite angles)
∴ By ASA rule ΔAPN ≅ ΔAQL
Hence, by CPCT i.e Corresponding parts of congruent triangles PA=AQ
Hence, A is equidistant from LM and KN.
The equation is already solved for you basically
Answer:
C
Step-by-step explanation:
Given
24x² - 22x + 5
Consider the factors of the product of the x² term and the constant term which sum to give the coefficient of the x- term.
product = 24 × 5 = 120 and sum = - 22
The factors are - 12 and - 10
Use these factors to split the x- term
24x² - 12x - 10x + 5 ( factor the first/second and third/fourth terms )
= 12x(2x - 1) + 5(2x - 1) ← factor out (2x - 1) from each term
= (12x - 5)(2x - 1) ← in factored form → C
Answer:
y - 1 = - 4(x + 3)
Step-by-step explanation:
The equation of a line in point- slope form is
y - b = m(x - a)
where m is the slope and (a, b) a point on the line
Here m = - 4 and (a, b) = (- 3, 1), hence
y - 1 = - 4(x - (- 3)), thus
y - 1 = - 4(x + 3) ← in point- slope form
