2x2-3x-20=0
4-3x-20=0
2-20=-16
-3x-16=0
-3x=16
x= -5.33333
I think i did it right. i hope this helps.
3x-5x+3 Answer:
Step-by-step explanation:
Ah yes, math, I believe the answer you are looking for, my young scholar, is A
To find the difference in weights from two years ago, you will need to combine the 2 differences from the two years.
-1.56 and 0.73 becomes -1.56 + 0.73.
To add a positive and a negative number you will subtract the absolute values of the numbers AND use the sign of the number with the bigger absolute value (negative) in your answer.
|-1.56| - |0.73|
1.56 - 0.73 = 0.83
-0.83
The cat lost 0.83 (-0.83) pounds over 2 years.
In order to solve for this question, let's assign a couple variables.
The variable 't' will represent the number of two-point questions, and the variable 's' will represent the number of six point questions.
From the given, we can already form two equations:
t + s = 36
("An exam... contains 36 questions")
2t + 6s = 148
("An exam worth 148 points... Some questions are worth 2 points, and the others are worth 6 points")
Before we begin calculating anything, we can simplify the second equation we made, since all the numbers are divisible by 2:
2t + 6s = 148
t + 3s = 74
Now let's refer back to the first equation. We can subtract both sides by 's' (you could also subtract both sides by 't', but I personally think that this will make solving the equation less difficult):
t + s = 36
t = 36 - s
This effectively gives us a value for the variable 't'. We can assign this value back into our first equation:
t + 3s = 74
(36 - s) + 3s = 74
36 + 2s = 74
2s = 38
s = 19
Input 's' into our second equation to solve for 't':
t = 36 - s
t = 36 - 19
t = 17
There are 17 two-point questions and 19 six-point questions
- T.B.
