If the equation is 6 = -10 - p/5 then multiply all terms by 5 :-
30 = -50 - p : p = -50 - 30 = -80
It is an equation with the plus or minus signs missing which makes it impossible to solve.
10p = 180 p = 18
10p + 5p - p = 15p - p = 14p
I think the answer is 8 but im not sure!02-10p plus 162-10= -8-8 + 16= 8
5p-15 = 15-5p 5p+5p = 15+15 10p = 30 p = 3 Note that the + and - sign changes when terms are moved over to the other side of the equation and that the solution is found by dividing each side by 10.
It is an equation with the plus or minus signs missing which makes it impossible to solve.
10p = 180 p = 18
-- Add 5p to each side of the equation. -- Add 15 to each side of the equation. -- Look at the equation and find the one last step that will give you " p = a number ", which is the solution.
10p + 5p - p = 15p - p = 14p
20
I think the answer is 8 but im not sure!02-10p plus 162-10= -8-8 + 16= 8
2p + 3q = 13, 5p - 4q = -2 Multiply the first equation by 4 and the second by 3 and add them, which gets rid of the q: 8p + 15p = 52 - 6, and 23p = 46, so p=2. Plug that into the first equation to find q: 4 + 3q = 13, so q=3. Test your answers in the second equation to be sure: 5(2) - 4(3) = 10-12 = -2. It checks. So p=2, q=3.
5p-15 = 15-5p 5p+5p = 15+15 10p = 30 p = 3 Note that the + and - sign changes when terms are moved over to the other side of the equation and that the solution is found by dividing each side by 10.
allow me to answer your question with another question: why is this in the calculus section?
the product of 10p (p–q) is 10p²-10pq Given: 10p (p–q) To find : the product of 10p (p–q) Solution: we have to find the product of 10p (p–q). so product of any number means the multiplication multiply (p–q). by 10p we get, =10p× (p–q) =10p×p-10p× q =10p²-10pq the product of 10p (p–q) is 10p²-10pq
This can be simplified to -10p = -150 p = -150/-10 p = 15
To make a pound using only 50p, 20p, and 10p coins, we can set up a linear Diophantine equation: 50x + 20y + 10z = 100, where x, y, and z are the number of each type of coin used. We can then use techniques such as generating functions or modular arithmetic to solve this equation and find the number of ways to make a pound. The solution will involve finding the integer solutions to the equation within a certain range, as there are constraints on the number of each type of coin available.