Let y = 3^x , then 9^x = (3^2)^x = (3^x)^2 = y^2, so we can write the equation as
y^2 - 10y + 9 = 0
(y - 9)(y - 1) = 0
giving y = 9 or y = 1
Therefore 3^x = 9 giving x = 2
and 3^x = 1 giving x = 0
so the solution set is {0, 2}
Note - I have assumed that the 'x' after the 10 means 'multiply'. Is that correct?
The cross-section cut by the plane yields a quadrilateral with length and width equal to the length and width of the face that is parallel to the plane.
Answer:
7 balls
Step-by-step explanation:
Look for the worst case scenario:
Get 3 white, 2 green, and 1 black.
That adds up to: 3 + 2 + 1 = 6.
+1 red ball: 6 + 1 =7
7 balls
F(x) = x^2 - 6x + 4 {-3, 0, 5} (x, y) x is the domain, y is the range.
plug in the domain numbers into the equation to find the range.
y = x^2 - 6x + 4
y = -3^2 - 6(-3) + 4
y = 9 - (-18) + 4
y = 9 + 18 + 4
y = 31 (-3, 31)
y = x^2 - 6x + 4
y = 0^2 - 6(0) + 4
y = 0 - 0 + 4
y = 4 (0, 4)
y = x^2 - 6x + 4
y = 5^2 - 6(5) + 4
y = 25 - 30 + 4
y = -1 (5, -1)
c. {-1, 4, 31}
hope this helped, God bless!
The answer to the question
