P(t+1) = -2
divide by P
Subtract 1
T equals negative 2 over P minus 1
T = -2/P -1
Answer:
56
Step-by-step explanation:
El problema se puede transcribir en esta ecuación:
2x + x = 168
siendo x las nectarinas
sumas los términos de x:
3x=168
despejas x:
x = 168 ÷ 3
x = 56
Answer:
give or take 5 miles.
Step-by-step explanation:
im not shor but that is true about the both of them
Answer:
2np + p²
Step-by-step explanation:
The general formula for the area of a square is A = s², where s = the length of one side of the square. In the case of the smaller square the area would be: n x n = n². Since the side of the larger square is 'p' inches longer, the length of one side is 'n + p'. To find the area of the larger square, we have to take the length x length or (n +p)².
Using FOIL (forward, outside, inside, last):
(n + p)(n+p) = n² + 2np + p²
Since the area of the first triangle is n², we can subtract this amount from the area of the larger square to find out how many square inches greater the larger square area is.
n² + 2np + p² - n² = 2np + p²
Answer: ![v=\sqrt[]{\frac{2K}{m} }](https://tex.z-dn.net/?f=v%3D%5Csqrt%5B%5D%7B%5Cfrac%7B2K%7D%7Bm%7D%20%7D)
Step-by-step explanation:
First, multiply by 2 to get rid of the 2 in the denominator. Remember that if you make any changes you have to make sure the equation keeps balanced, so do it on both sides as following;
Divide by m to isolate 
.
To eliminate the square and isolate v, extract the square root.
![\sqrt[]{\frac{2K}{m} }=\sqrt[]{v^2}](https://tex.z-dn.net/?f=%5Csqrt%5B%5D%7B%5Cfrac%7B2K%7D%7Bm%7D%20%7D%3D%5Csqrt%5B%5D%7Bv%5E2%7D)
![\sqrt[]{\frac{2K}{m} }=v](https://tex.z-dn.net/?f=%5Csqrt%5B%5D%7B%5Cfrac%7B2K%7D%7Bm%7D%20%7D%3Dv)
let's rewrite it in a way that v is in the left side.
