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
24 half-kg packets = 24 x 1/2 = 12kg
Sugar remained = 50 - 12 = 38 kg
Answer: 38 kg
Answer:
Five touchdowns were made in the game
Step-by-step explanation:
Here, we want to know the number of touchdowns made during the game.
We proceed as follows;
Let the number of touch downs be x
So the total points earned through touchdowns is 6 * x = 6x
The number of scores is 10 times
Let the number of extra kick be y which means that the number of conversions will be (y-1)
So the total number of score times will be ;
x + y + y-1 = 10
x + 2y = 11 •••••••••(i)
Now let’s work with points
Touch down points = 6 * x = 6x
Points from extra kick = 1 * y = y
Points from 2-point conversions = 2(y-1) = 2y - 2
So;
6x + y + 2y -2 = 37
6x + 3y = 37 + 2
6x + 3y = 39
divide through by 3
2x + y = 13 •••••••(ii)
So now solve simultaneously
From ii, y = 13 - 2x
Put this into i
x + 2(13-2x) = 11
x + 26 - 4x = 11
x -4x = 11-26
-3x = -15
x = -15/-3
x = 5
There are five touchdowns in the game
Answer:
7y+24
Step-by-step explanation:
