Consecutive integers are 1 apart
x,x+1,x+2
(x)(x+1)(x+2)=-120
x^3+3x^2+3x=-120
add 120 to both sides
x^3+3x^2+3x+120=0
factor
(x+6)(x^2-3x+20)=0
set each to zero
x+6=0
x=-6
x^2-3x+20=0
will yeild non-real result, discard
x=-6
x+1=-5
x+2=-4
the numbers are -4,-5,-6
use trial and error and logic
factor 120
120=2*2*2*3*5
how can we rearange these numbers in (x)(y)(z) format such that they multiply to 120?
obviously, the 5 has to stay since 2*5=10 which is out of range
so 2*2*2*3 has to arrange to get 3,4 or 4, 6 or 6,7
obviously, 7 cannot happen since it is prime
3 and 4 results in in 12, but 2*2*2*3=24
therfor answer is 4 and 6
they are all negative since negaive cancel except 1
the numbers are -4,-5,-6
If j=23 and j-5 is how old Jeffery's brother is then Jeffrey's brother is 17 years old.
So j=17
Find the area of the larger rectangle.
5 x 4 = 20
One of the triangles is 2 x 2. Area of 2. Subtract 2 from 20.
20 - 2 = 18
Another is 2 x 5. Area of 5. Subtract 5 from 18.
18 - 5 = 13
The last is 3 x 4. Area of 6. Subtract 6 from 13.
13 - 6 = 7
The area of the triangle QRS is A, 7.
Hope this helps!
The Answer is : (x - 3)/(x + 2) not x+2/x+3 Thus A) is your Answer
Simplify the following:
((x^2 + x - 6) (x^2 - 9))/((x^2 - 4) (x^2 + 6 x + 9))
The factors of -6 that sum to are 3 and -2. So, x^2 + x - 6 = (x + 3) (x - 2):
((x + 3) (x - 2) (x^2 - 9))/((x^2 - 4) (x^2 + 6 x + 9))
The factors of 9 that sum to 6 are 3 and 3. So, x^2 + 6 x + 9 = (x + 3) (x + 3):
((x + 3) (x - 2) (x^2 - 9))/((x + 3) (x + 3) (x^2 - 4))
(x + 3) (x + 3) = (x + 3)^2:
((x + 3) (x - 2) (x^2 - 9))/((x + 3)^2 (x^2 - 4))
x^2 - 4 = x^2 - 2^2:
((x + 3) (x - 2) (x^2 - 9))/((x^2 - 2^2) (x + 3)^2)
Factor the difference of two squares. x^2 - 2^2 = (x - 2) (x + 2):
((x + 3) (x - 2) (x^2 - 9))/((x - 2) (x + 2) (x + 3)^2)
x^2 - 9 = x^2 - 3^2:
((x + 3) (x - 2) (x^2 - 3^2))/((x - 2) (x + 2) (x + 3)^2)
Factor the difference of two squares. x^2 - 3^2 = (x - 3) (x + 3):
((x - 3) (x + 3) (x + 3) (x - 2))/((x - 2) (x + 2) (x + 3)^2)
((x + 3) (x - 2) (x - 3) (x + 3))/((x - 2) (x + 2) (x + 3)^2) = (x - 2)/(x - 2)×((x + 3) (x - 3) (x + 3))/((x + 2) (x + 3)^2) = ((x + 3) (x - 3) (x + 3))/((x + 2) (x + 3)^2):
((x + 3) (x - 3) (x + 3))/((x + 2) (x + 3)^2)
Combine powers. ((x + 3) (x - 3) (x + 3))/((x + 2) (x + 3)^2) = ((x + 3)^(1 + 1) (x - 3))/((x + 2) (x + 3)^2):
((x + 3)^(1 + 1) (x - 3))/((x + 3)^2 (x + 2))
1 + 1 = 2:
((x + 3)^2 (x - 3))/((x + 2) (x + 3)^2)
Cancel terms. ((x + 3)^2 (x - 3))/((x + 2) (x + 3)^2) = (x - 3)/(x + 2):
Answer: (x - 3)/(x + 2)
The probability of getting heads on the first toss is 1/2 .
Then, the probability of getting heads on the second toss is 1/2 .
Then, the probability of getting heads on the third toss is 1/2 .
The probability of all three things happening is ...
(1/2) x (1/2) x (1/2) = 1/8 = 0.125 = 12.5% .
