Simplify the following:
((x^2 - 11 x + 30) (x^2 + 6 x + 5))/((x^2 - 25) (x - 5 x - 6))
The factors of 5 that sum to 6 are 5 and 1. So, x^2 + 6 x + 5 = (x + 5) (x + 1):
((x + 5) (x + 1) (x^2 - 11 x + 30))/((x^2 - 25) (x - 5 x - 6))
The factors of 30 that sum to -11 are -5 and -6. So, x^2 - 11 x + 30 = (x - 5) (x - 6):
((x - 5) (x - 6) (x + 5) (x + 1))/((x^2 - 25) (x - 5 x - 6))
x - 5 x = -4 x:
((x - 5) (x - 6) (x + 5) (x + 1))/((x^2 - 25) (-4 x - 6))
Factor -2 out of -4 x - 6:
((x - 5) (x - 6) (x + 5) (x + 1))/(-2 (2 x + 3) (x^2 - 25))
x^2 - 25 = x^2 - 5^2:
((x - 5) (x - 6) (x + 5) (x + 1))/(-2 (x^2 - 5^2) (2 x + 3))
Factor the difference of two squares. x^2 - 5^2 = (x - 5) (x + 5):
((x - 5) (x - 6) (x + 5) (x + 1))/(-2(x - 5) (x + 5) (2 x + 3))
((x - 5) (x - 6) (x + 5) (x + 1))/((x - 5) (x + 5) (-2) (2 x + 3)) = ((x - 5) (x + 5))/((x - 5) (x + 5))×((x - 6) (x + 1))/(-2 (2 x + 3)) = ((x - 6) (x + 1))/(-2 (2 x + 3)):
((x - 6) (x + 1))/(-2 (2 x + 3))
Multiply numerator and denominator of ((x - 6) (x + 1))/(-2 (2 x + 3)) by -1:
Answer: (-(x - 6) (x + 1))/(2 (2 x + 3))
I think it’s the red line.
So you want to put the equation into y=mx+b so you would need to subtract the x to put it on the right side so then you’ll have 2y=-x+0 then divide by 2 to get the y by itself then you’ll have y=-2x+0 that means the line would be negative and the y intercept is 0 and the only line that follows that is the red line.
Hope this helped :)
A. √(0.8^2) + (0.6^2) = √1 = 1 => OK
<span>b.(-2/3,√ 5/3) = √(-2/3)^2 + 5/9) = √(4/9 +5/9) = √1 = 1 => OK
c.(√ 3/2, 1/3) = √(3/4 + 1/9) < 1 => it is inside the unit circle
d.(1,1)
= √(1 + 1) = √2 > 1 => NO. This point is beyond the limits of the unit circle.</span>
Written just as an expression it would be:
15-5(p-6)
Answer:
C
Step-by-step explanation:
10^5=100,000
100,000*4.7=470,000 (Four Hundred Seventy Thousand)
10^8=100000000
100000000*3.9=390,000,000 (Three Hundred Ninty Million)
10^5=100,000
100,000*5.2=520,000 (Five Hundred Two Thousand)
Least to Greatest:
470,000 520,000 390,000,000
I III II
C
