Firstly expand (x - 3)(ax^2 + bx + c):
ax^3 + bx^2 + cx - 3ax^2 - 3bx - 3c
Now rearrange it so that it's in the form ax^3 + bx^2 + cx + d:
ax^3 + (b-3a)x^2 + (c-3b)x - 3c
Now we can compare both equations:
ax^3 + (b-3a)x^2 + (c-3b)x - 3c = 2x^3 - x^2 - 19x + 12
We get:
(1) a = 2
(2) b - 3a = -1
(3) c - 3b = -19
(4) -3c = 12
If we substitute (1) into (2) we get:
b - 3*2 = -1
b - 6 = -1
b = 5
Now if we solve (4) we get:
-3c = 12
c = -4
Therefor a = 2, b = 5 and c = -4
Answer:
v=0
Step-by-step explanation:
I think you mean 7(2+5v)=2v+14.
14+35v=2v+14 distribute
33v=0 Subtract 2v and 14
v=0 divide by 33 on both sides
Hope this helps plz mark brainliest if correct :D
At 8 its 3 below 0 and at 10 its 1 above 0 so its 4° different
Answer:
The answer is 1,000
SInce the beginning number is 3 and there are ten possible numbers to put in the remaining three slots, there are exactly 1,000 possible combinations for a 3-digit code. The answer is 1,000. There are 3 rows of 10 digits. The number of combinations 10 to the thid power which is 1000 (10 * 10 * 10)
