Answer:
6x^3(x+2)(x-7)
Step-by-step explanation:
1st step: factor out common term, 6x^3(x^2 - 5x - 14)
2nd step: Factor x^2 - 5x - 14, (x + 2)(x - 7)
3rd step: simplify, 6x^3(x+2)(x-7)
I am joyus to assist :)
Answer:
42 * 10 + 42 * 5
which agrees with the third option (C) shown on the list of possible answers
Step-by-step explanation:
Notice that using distributive property we can write:
42 * 15 = 42 (10 + 5) = 42 * 10 + 42 * 5
which agrees with the third option (C) shown on the list of possible answers.
Answer:
f⁻¹(x) = (1/2)x +5
Step-by-step explanation:
In y = f(x), swap the variables, then solve for y. The expression you get is f⁻¹(x).
... y = 2x -10
... x = 2y -10 . . . . . . swapped variables
... x +10 = 2y . . . . . add 10
... (1/2)x + 5 = y . . . . divide by 2
... f⁻¹(x) = (1/2)x + 5 . . . . . . rewrite using function notation
Answer:
The population of bacteria can be expressed as a function of number of days.
Population = ![\[5*2^{n-1}\]](https://tex.z-dn.net/?f=%5C%5B5%2A2%5E%7Bn-1%7D%5C%5D)
where n is the number of days since the beginning.
Step-by-step explanation:
Number of bacteria on the first day=![\[5 * 2^{0} = 5\]](https://tex.z-dn.net/?f=%5C%5B5%20%2A%202%5E%7B0%7D%20%3D%205%5C%5D)
Number of bacteria on the second day = ![\[5 * 2^{1} = 10\]](https://tex.z-dn.net/?f=%5C%5B5%20%2A%202%5E%7B1%7D%20%3D%2010%5C%5D)
Number of bacteria on the third day = ![\[5*2^{2} = 20\]](https://tex.z-dn.net/?f=%5C%5B5%2A2%5E%7B2%7D%20%3D%2020%5C%5D)
Number of bacteria on the fourth day = ![\[5*2^{3} = 40\]](https://tex.z-dn.net/?f=%5C%5B5%2A2%5E%7B3%7D%20%3D%2040%5C%5D)
As we can see , the number of bacteria on any given day is a function of the number of days n.
This expression can be expressed generally as where n is the number of days since the beginning.
Answer:
I think A is the answer
but D is the correct answer "kunno"
Step-by-step explanation:
hope it helps :)
