Answer:
1/2
Step-by-step explanation:
The unit fraction for an area (or anything, for that matter) divided into n equal parts is 1/n. For 2 equal parts, it is 1/2.
Answer:
well duh Mr.Sanchez's class
Step-by-step explanation:
so the points are, from P1 to P2, namely P1P2, and from P2 to P3, namely P2P3, and from P3 back to P1, namely P3P1.
![\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ P1(\stackrel{x_1}{5}~,~\stackrel{y_1}{-4})\qquad P2(\stackrel{x_2}{8}~,~\stackrel{y_2}{-3})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ P1P2=\sqrt{[8-5]^2+[-3-(-4)]^2}\implies P1P2=\sqrt{(8-5)^2+(-3+4)^2} \\\\\\ P1P2=\sqrt{3^2+1^2}\implies \boxed{P1P2=\sqrt{10}}\\\\[-0.35em] \rule{34em}{0.25pt}](https://tex.z-dn.net/?f=%20%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%20%5C%5C%5C%5C%20P1%28%5Cstackrel%7Bx_1%7D%7B5%7D~%2C~%5Cstackrel%7By_1%7D%7B-4%7D%29%5Cqquad%20%20P2%28%5Cstackrel%7Bx_2%7D%7B8%7D~%2C~%5Cstackrel%7By_2%7D%7B-3%7D%29%5Cqquad%20%5Cqquad%20%20d%20%3D%20%5Csqrt%7B%28%20x_2-%20x_1%29%5E2%20%2B%20%28%20y_2-%20y_1%29%5E2%7D%20%5C%5C%5C%5C%5C%5C%20P1P2%3D%5Csqrt%7B%5B8-5%5D%5E2%2B%5B-3-%28-4%29%5D%5E2%7D%5Cimplies%20P1P2%3D%5Csqrt%7B%288-5%29%5E2%2B%28-3%2B4%29%5E2%7D%20%5C%5C%5C%5C%5C%5C%20P1P2%3D%5Csqrt%7B3%5E2%2B1%5E2%7D%5Cimplies%20%5Cboxed%7BP1P2%3D%5Csqrt%7B10%7D%7D%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%20)
![\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ P2(\stackrel{x_2}{8}~,~\stackrel{y_2}{-3})\qquad P3(\stackrel{x_2}{7}~,~\stackrel{y_2}{-10}) \\\\\\ P2P3=\sqrt{[7-8]^2+[-10-(-3)]^2}\implies P2P3=\sqrt{(7-8)^2+(-10+3)^2} \\\\\\ P2P3=\sqrt{(-1)^2+(-7)^2}\implies P2P3=\sqrt{50}\implies \boxed{P2P3=5\sqrt{2}}\\\\[-0.35em] \rule{34em}{0.25pt}](https://tex.z-dn.net/?f=%20%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%20%5C%5C%5C%5C%20P2%28%5Cstackrel%7Bx_2%7D%7B8%7D~%2C~%5Cstackrel%7By_2%7D%7B-3%7D%29%5Cqquad%20%20P3%28%5Cstackrel%7Bx_2%7D%7B7%7D~%2C~%5Cstackrel%7By_2%7D%7B-10%7D%29%20%5C%5C%5C%5C%5C%5C%20P2P3%3D%5Csqrt%7B%5B7-8%5D%5E2%2B%5B-10-%28-3%29%5D%5E2%7D%5Cimplies%20P2P3%3D%5Csqrt%7B%287-8%29%5E2%2B%28-10%2B3%29%5E2%7D%20%5C%5C%5C%5C%5C%5C%20P2P3%3D%5Csqrt%7B%28-1%29%5E2%2B%28-7%29%5E2%7D%5Cimplies%20P2P3%3D%5Csqrt%7B50%7D%5Cimplies%20%5Cboxed%7BP2P3%3D5%5Csqrt%7B2%7D%7D%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%20)
![\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ P3(\stackrel{x_2}{7}~,~\stackrel{y_2}{-10})\qquad P1(\stackrel{x_1}{5}~,~\stackrel{y_1}{-4}) \\\\\\ P3P1=\sqrt{[5-7]^2+[-4-(-10)]^2}\implies P3P1=\sqrt{(5-7)^2+(-4+10)^2} \\\\\\ P3P1=\sqrt{(-2)^2+6^2}\implies P3P1=\sqrt{40}\implies \boxed{P3P1=2\sqrt{10}}](https://tex.z-dn.net/?f=%20%5Cbf%20~~~~~~~~~~~~%5Ctextit%7Bdistance%20between%202%20points%7D%20%5C%5C%5C%5C%20P3%28%5Cstackrel%7Bx_2%7D%7B7%7D~%2C~%5Cstackrel%7By_2%7D%7B-10%7D%29%5Cqquad%20%20P1%28%5Cstackrel%7Bx_1%7D%7B5%7D~%2C~%5Cstackrel%7By_1%7D%7B-4%7D%29%20%5C%5C%5C%5C%5C%5C%20P3P1%3D%5Csqrt%7B%5B5-7%5D%5E2%2B%5B-4-%28-10%29%5D%5E2%7D%5Cimplies%20P3P1%3D%5Csqrt%7B%285-7%29%5E2%2B%28-4%2B10%29%5E2%7D%20%5C%5C%5C%5C%5C%5C%20P3P1%3D%5Csqrt%7B%28-2%29%5E2%2B6%5E2%7D%5Cimplies%20P3P1%3D%5Csqrt%7B40%7D%5Cimplies%20%5Cboxed%7BP3P1%3D2%5Csqrt%7B10%7D%7D%20)
Answer:
For the function V(t)=24300(1.37)t, the rate of increase is 37%.
Step-by-step explanation:
The function represents the value (V) of the car over time (t). This type of function is exponential growth, which means for each year, the value of the vintage car will increase by a rate of 37%. Exponential growth functions are represented by the equation f(x)=ab^x, where 'a'=initial value, 'b'=the rate and 'x' represents time. In this case, our initial value of the car is $24,300 and the rate is 1.37. A rate of 1.37 indicates that the car will retain its initial value (1) as well as increase be an additional 37 percent (.37) over time.
To answer the question above, determine the velocity of the car at time zero. To do such, determine the increase in velocity every hour (acceleration).
a = (62 - 53) / (6 -3) = 3 km/ h^2
Calculate for the initial velocity (v) by using either of the velocities paired with the corresponding time,
v = 53 km/h - (3 km/h^2) x 3 h = 44 km/h
The equation is <em>y = 44 + 3x</em>. Where x is the time in hours, y is the velocity in km/h.
