Answer:
1/3^2
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The domain is a discrete set. So we will have a discrete range.
The range is a set containing:
f(-2)=-11
f(-1)=-9
f(0) = -7
f(1) = -5
f(2) = -3
Range = {-11,-9,-7,-5,-3}
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! :)
Answer:
x = 2.5 or 7.5
Step-by-step explanation:
Subtract 3/4.
|5 -x| = 10/4 = 2.5
This resolves to two equations:
- 5 -x = 2.5 ⇒ x = 5 -2.5 = 2.5
- 5 -x = -2.5 ⇒ x = 5 +2.5 = 7.5
The solutions are x = 2.5 or 7.5.
