The answer
let be
A1 = (2x^4-3x^3+2x+4)-(-x^4-3x^3+4x^2+x+3) =
= 2x^4-3x^3+2x+4 +x^4 +3x^3- 4x^2- x-3
A1= 6x^4 -2x² +x +1
and
A2= 5x^4-2x^3+3x^2+3x-4-(2x^4-2x^3+7x^2+2x-5)
= 5x^4-2x^3+3x^2+3x-4-2x^4+2x^3-7x^2-2x+5
A2<span>= 3x^4-4x²+x+1
A1 is not equivalent to A2
so the answer is </span><span>B. False </span>
The last two options are correct, but it looks like you’ve got them :)
1. <
2. >
3. <
4. >
these are the right answers. i just plugged the numbers into my calculator:)
Combine like terms in Qx + 12 = 13x + P:
x(Q-13) = P-12
P-12
Then x = ----------
Q-13
Note that Q may not = 13. But none of the answer choices present Q = 13.
Let's go thru the answer choices one by one.
P-12
x = ----------
Q-13
-24
Check out A: Q=12 and P= -12: x = -------- = 24 This is OK (ONE sol'n)
-1
-25
Check out B: Q = -13 and P = -13: x = ----------- = 25/26 OK
-26
13-12
Check out C: Q = -13 and P = 13: x = ------------ = -1/26 OK
-13-13
12-12
Check out D: Q = 12 and P = 12: x= ---------- = 0 OK
12-13
It appears that in all four cases, the equation has ONE solution.
Does the question start with how many were in the bag originally?