X ^2 +6x+ 5
(x+2)(x+3)
2 + 3 = 5
While also 2•3= 6
The two numbers have to equal b when multiplied and c when added
Answer:
P and Q are two points on the line x-y+1=0 and are at a distant of 5 units from the origin. Find the area of triangle POQ.
Step-by-step explanation:
P and Q are the intersection points of
x-y+1 = 0 and the circle x^2 + y^2 = 25
sub y = x+1 into the circle
x^2 + (x+1)^2 = 25
x^2 + x^2 + 2x + 1 - 25 = 0
x^2 + x - 12 = 0
(x+4)(x-3) = 0
x = 3 or x = -4
y = 4 or y = -3
so P(3,4) and Q(-4,3) are our two points
Height of triangle.
h = |0 - 0 + 1|/√2 = 1/√2
PQ = √( (-7)^2 + 1^2) = √50 = 5√2
area POQ = (1/2)(1/√2)(5√2) = 5/2 square units
hope this helped
Answer:
19y-9
Step-by-step explanation:
apply distributive property
-3(-4y)-3 x 3 + 7y
12y - 3 x 3 + 7y
12y - 9 + 7y
add 12y to 7y
19y-9
