P2 + 13p - 30 = 0 Answer: p= -15, p = 2
It depends on what aspect of constant velocity you are talking about. Since the velocity is not changing, one valid equation is: V = [number] At the same time, acceleration is zero, so another equation is: A = 0 If "p" is position and p1 is the original position and p2 is the current position after tine lapse "t," then: p2 = p1 + Vt
p2 + 2pq + q2 = 1q2 + 2pq + (p2 - 1) = 0q = 1/2 [ -2p plus or minus sqrt( 4p2 - 4p2 + 4 ) ]q = -1 - pq = 1 - p
You use the well-known point-point formula . Here is it: suppose P1 = (x1,y1) and P2 = (x2,y2) An equation for the line containing P1 and P2 is y - y1 = [(y1-y2)/(x1-x2)] (x -x1) Note that the quantity in brackets is the slope of the line. Note also that it does not matter which point is P1 and which is P2.
Take any two points and form the equation for a straight line. If all the remaining points satisfy the equation, then they lie on astraight line. Else, they don't. Here's an example. Consider n points as P1(x1, y1), P2(x2, y2), ...., Pn(xn, yn). In order to determine if P1, P2, ..., Pn lie on a straight line, form the straight line equation with P1 and P2 as: y-y1= m * (x - x1), where the slope m = (y2-y1)/(x2-x1). Then try to satisfy this equation by the remaining points P3, P4, ..., Pn. That is, verify the following: Is y3-y1= m * (x3 - x1)? Is y4-y1= m * (x4 - x1)? ... Is yn-y1= m * (xn - x1)? If all of the above is true, then the points lie on a straight line.
The frequency of the homozygous dominant genotype.
To solve Boyle's Law equation for V2, first write the equation as P1V1 = P2V2. Then rearrange it to isolate V2 on one side, dividing both sides by P2 to solve for V2, which will be V2 = (P1 * V1) / P2.
P2 + 13p - 30 = 0 Answer: p= -15, p = 2
It depends on what aspect of constant velocity you are talking about. Since the velocity is not changing, one valid equation is: V = [number] At the same time, acceleration is zero, so another equation is: A = 0 If "p" is position and p1 is the original position and p2 is the current position after tine lapse "t," then: p2 = p1 + Vt
p2 + 2pq + q2 = 1q2 + 2pq + (p2 - 1) = 0q = 1/2 [ -2p plus or minus sqrt( 4p2 - 4p2 + 4 ) ]q = -1 - pq = 1 - p
You use the well-known point-point formula . Here is it: suppose P1 = (x1,y1) and P2 = (x2,y2) An equation for the line containing P1 and P2 is y - y1 = [(y1-y2)/(x1-x2)] (x -x1) Note that the quantity in brackets is the slope of the line. Note also that it does not matter which point is P1 and which is P2.
Using the combined gas law equation (P1V1 = P2V2), we can find the new pressure. Rearranging the equation to solve for P2 gives us P2 = (P1V1) / V2. Plugging in the values gives P2 = (74 atm * 190 mL) / 30.0 mL = 470 atm. Therefore, the gas's new pressure is 470 atm.
If you have p1, p2, p3, and p4, it seems like you are referring to four different entities, possibly variables or parameters in a system or mathematical equation. Without further context, it is difficult to provide a specific interpretation or action related to p1, p2, p3, and p4.
(p1/v1) = (p2/v2)For Apex (P1 N1)= (P2N2 )
It is not an equation, but q2 meaning q^2 represents q being multiplied by itself.
If you dcrease the temperature you will decrease the pressure proportionately. So, T1 over T2 will equal P1 over P2. We can derive the formula P1 x T2 = P2 x T1. Substitue the values and we get 50.0 mm x 200K = P2 x 540K 10,000 mmK = P2 x 540K 10,000mmK / 540K = P2 P2 = 18.52 mm of Mercury in a constant volume
Pascal's law states that a change in pressure applied to an enclosed fluid is transmitted undiminished to all portions of the fluid and to the walls of its container. The equation for Pascal's law is P1 = P2, where P1 is the initial pressure applied to the fluid and P2 is the final pressure exerted throughout the fluid.