Let's solve the equation 2k^2 = 9 + 3k
First, subtract each side by (9+3k) to get 0 on the right side of the equation
2k^2 = 9 + 3k
2k^2 - (9+3k) = 9+3k - (9+3k)
2k^2 - 9 - 3k = 9 + 3k - 9 - 3k
2k^2 - 3k - 9 = 0
As you see, we got a quadratic equation of general form ax^2 + bx + c, in which a = 2, b= -3, and c = -9.
Δ = b^2 - 4ac
Δ = (-3)^2 - 4 (2)(-9)
Δ<u /> = 9 + 72
Δ<u /> = 81
Δ<u />>0 so the equation got 2 real solutions:
k = (-b + √Δ)/2a = (-(-3) + √<u />81) / 2*2 = (3+9)/4 = 12/4 = 3
AND
k = (-b -√Δ)/2a = (-(-3) - √<u />81)/2*2 = (3-9)/4 = -6/4 = -3/2
So the solutions to 2k^2 = 9+3k are k=3 and k=-3/2
A rational number is either an integer number, or a decimal number that got a definitive number of digits after the decimal point.
3 is an integer number, so it's rational.
-3/2 = -1.5, and -1.5 got a definitive number of digit after the decimal point, so it's rational.
So 2k^2 = 9 + 3k have two rational solutions (Option B).
Hope this Helps! :)
Pretty sure it's A. (X times .5 equals Y)
If you graphed the equation: y = 5
You would put a straight line going horizontally through 5. Like this picture shows. If it were x = 5, for example, then it would go up and down through 5.
Let me know if this helped! Have a nice day!
Answer:
x = 3 ±sqrt(6)
Step-by-step explanation:
6(x-3)^2-26 = 10
Add 26 to each side
6(x-3)^2-26+26 = 10+26
6(x-3)^2 = 36
Divide by 6
6/6(x-3)^2 = 36/6
(x-3)^2 = 6
Take the square root of each side
sqrt((x-3)^2) = ±sqrt(6)
x-3 = ±sqrt(6)
Add 3 to each side
x-3+3 =3 ±sqrt(6)
x = 3 ±sqrt(6)
Answer:
Since he earned one third as many as Jose
it's 8
Step-by-step explanation:
