Find factors of -5 that, when added together, will give 4
x² + 4x - 5
x 5
x -1
(x + 5)(x -1)
Using the FOIL method (First, Outside, Inside, Last), we can get the question again.
x(x) = x²
x(-1) = -x
5(x) = 5x
5(-1) = -5
x² - x + 5x - 5
x² + 4x - 5
(x + 5) (x - 1) is your answers
however, i believe it is only asking for one. Therefore, because (x - 1) is a choice, (x - 1) should be your answer
hope this helps
Answer:
Step-by-step explanation:
1 × 5 = 5
(- 2) × (- 3) = 6
4 × (- 6) = - 24
I hope I've helped you.
Answer:
(4,1)
Step-by-step explanation:
let J(6,6) be x1, y1
let K(2,-4) be x2, y2
midpoint = (x1+x2)/2 , (y1+y2)/2
=(6+2)/2 ,(6-4)/2
= 8/2 ,2/2
=4, 1
Answer:
Proved See below
Step-by-step explanation:
Man this one is a world of its own :D Just a quick question are you a fellow Add Math student in O levels i remember this question from back in the day :D Anyhow Lets get started
For this question we need to know the following identities:

Lets solve the bottom most part first:

Take LCM

now break the LCM

because 1/tan = cot x
and furthermore,

now we solve the above part and replace the bottom most part that we solved with 

Hence proved! :D
Answer:
x = 25, y = 9
Step-by-step explanation:
Since the triangles are congruent then corresponding angles and sides are congruent, thus
∠ O = ∠ P , substitute values
6y - 14 = 40 ( add 14 to both sides )
6y = 54 ( divide both sides by 6 )
y = 9
and
NG = IT , that is
x - 2y = 7 ( substitute y = 9 into the equation )
x - 2(9) = 7
x - 18 = 7 ( add 18 to both sides )
x = 25