Let x and y be the 2 parts of 15 ==> x + y=15 (given)
Reciprocal of x and y ==> 1/x +1/y ==> 1/x + 1/y = 3/10 (given)
Let's solve 1/x + 1/y = 3/10 . Common denominator = 10.x.y (reduce to same denominator)
==> (10y+10x)/10xy = 3xy/10xy ==> 10x+10y =3xy
But x+y = 15 , then 10x+10y =150 ==> 150=3xy and xy = 50
Now we have the sum S of the 2 parts that is S = 15 and
their Product = xy =50
Let's use the quadratic equation for S and P==> X² -SX +P =0
Or X² - 15X + 50=0, Solve for X & you will find:
The 1st part of 15 is 10 & the 2nd part is 5
P(black socks): 24/42 or 12/21
P(black socks without replacing): 23/41. As a result, the probability of randomly picking 2 black socks, without replacement, from the basket is 12/21×23/41=276/861 or 32%. Hope it help!
Answer:
x = ±
- 3
Explanation:
I'm assuming you want the solutions to that equation, so here goes! (If not, please comment.)
(x-3)(x+9)=27
Let's FOIL this all out and expand. (Remember: First, Outer, Inner, Last.)
x^2 + 9x - 3x - 27
(first+ inner + outer + last)
x^2 + 9x - 3x - 27 = 27
Combine like terms, and add 27 to both sides.
x^2 + 6x - 27 + 27 = 27 + 27
x^2 + 6x = 54
Let's complete the square, because factoring doesn't work, and because it's good practice.
x^2 + 6x + ___ = 54 + ____
In the blank we will put b/2 ^2 = 6/2 ^2 = 3^2 = 9 to complete the square.
x^2 + 6x + 9 = 54 + 9
Now we've got a perfect square factor:
(x + 3)^2 = 63
sqrt(x+3)^2 = 
x + 3 = ± 
x = ±
- 3