We have:
(30x²+23x+16)/(cx+3) - 13/(cx+3) = 6x+1
(30x²+23x+16 - 13)/(cx+3) = 6x+1
(30x²+23x+3)/(cx+3) = 6x+1
30x²+23x+3 = (cx+3)(6x+1)
30x²+23x+3 = 6cx²+cx+18x+3
30x² + 23x + 3 - 6cx² - cx - 18x - 3 = 0
(30 - 6c)x² +(5 - c)x = 0
6(5 - c)x² +(5 - c)x = 0
(5 - c)(6x² +x) = 0, and x∈ R\ {3/c} ⇒ 5 - c = 0 ⇒ c = 5.
Answer:
(5 x + 2) (x + 2)
Step-by-step explanation:
Factor the following:
5 x^2 + 10 x + 2 x + 4
10 x + 2 x = 12 x:
5 x^2 + 12 x + 4
Factor the quadratic 5 x^2 + 12 x + 4. The coefficient of x^2 is 5 and the constant term is 4. The product of 5 and 4 is 20. The factors of 20 which sum to 12 are 2 and 10. So 5 x^2 + 12 x + 4 = 5 x^2 + 10 x + 2 x + 4 = 2 (5 x + 2) + x (5 x + 2):
2 (5 x + 2) + x (5 x + 2)
Factor 5 x + 2 from 2 (5 x + 2) + x (5 x + 2):
Answer: (5 x + 2) (x + 2)
It would be x=3.50 because 8 divided by 4 is 2 and 14 divided by 4 is 3.5
Answer:
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The coordinates of the image point are given to be B(4, -8). We are to find the coordinates of pre-image that is the coordinates before the translation.
The translation made was (x-2, y+3)
This means, the x coordinate was moved 2 units to left and y coordinate was moved 3 units above.
So, the coordinates of original point will be, 2 units to right of B and 3 units down of B.
So the coordinates will be (6, -11)
