allow me to answer your question with another question:
why is this in the calculus section?
4p - 12 = 3p - 8 - 9p 4p - 3p + 9p = -8 + 12 10p = 4 p = 0.4 (or 2/5)
9p + 12
-9p + 6p = -3p
-14 + 3p = -9p - 21 add 14 and 9p to both sides, then 12p = -7 p = -7/12
9p + 27 = 9(p + 3)
4p - 12 = 3p - 8 - 9p 4p - 3p + 9p = -8 + 12 10p = 4 p = 0.4 (or 2/5)
9p + 12
-9p + 6p = -3p
-14 + 3p = -9p - 21 add 14 and 9p to both sides, then 12p = -7 p = -7/12
9p + 27 = 9(p + 3)
The equation (9p + 8 = 10p + 7) is an open equation because it contains a variable, (p), and is not universally true or false for all values of (p). To determine if it can be true for some value of (p), we can rearrange it. Subtracting (9p) from both sides gives (8 = p + 7), which simplifies to (p = 1). Therefore, the equation is true when (p = 1), but false for other values of (p).
Evaluate 9p + 33 for p = 5
9p - 5
Hi I am in grade 7. The grade 7 standed is 9P. Grade7- 9P Grade8- 10P Grade9- 11P Grade10- 12P
-9p-17=8 -17=8+9p 9p=-17-8 9p= -25 p=-25/9
Equilibrium occurs where Qd = QsQd - 128 + 9p = 0Qd = 128 - 9pQs + 32 - 7p = 0Qs = 7p - 327p - 32 = 128 - 9p16p = 160p = 10
-10 -7 + 9p = 9p - 17