If in one hour Belinda can go 10 miles that means in 30 minutes she can go 5 miles. 60/2=30 10/2=5
If she can go 5 miles is 30 minutes than by doing some simple math she must be able to go 2.5 miles in 15 mins. 30/2=15 5/2=2.5
Meaning that when you add 30+15 you get 45 and when you add the distance she can go in each amount of given time 5+2.5 you get 7.5
- So the trail has to be 7.5 miles long
(i) speed = distance / time
so time = distance / speed
here we have
time t = 1080/x hours
(ii) return flight time = 1080 / (x + 30) hours
(a) 1080/x - 1080/(x + 30) = 1/2
Multiplying through by the LCD 2x(x + 30) we get:-
1080*2(x + 30) - 2x*1080 = x(x+30)
2160x + 64800 - 2160x = x^2 + 30x
x^2 + 30x - 64800 = 0
(b) factoring; -64800 = 270 * -240 ans 270-240 = 30 so we have
(x + 270)(x - 240) = 0 so x = 240 ( we ignore the negative -270)
So the speed for outward journey is 240 km/hr
(c) time ffor outward flight = 1080 / 240 = 4 1/2 hours
(d) average speed for whole flight = distance / time
Time for outward journey = 4.5 hours and time for return journey = d / v
= 1080 / (240+30) = 4 hours
Therefore the average speed for whole journey = 2160 / 8.5 = 254.1 km/hr
Okay doll so this is how you do it.
The 12 ounces doesn't matter. Its irrelevant.
You divide 14.22 by 18.
And that's all there is to it.
The answer is C. $0.79
Answer:
priority 1st for the equation inside brackets
so 11+-1
=10
let's keep in mind that perpendicular lines have <u>negative reciprocal</u> slopes, hmmm what is the slope of line S anyway?
![\bf (\stackrel{x_1}{2}~,~\stackrel{y_1}{7})\qquad (\stackrel{x_2}{6}~,~\stackrel{y_2}{1}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{1-7}{6-2}\implies \cfrac{-6}{4}\implies -\cfrac{3}{2} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{-\cfrac{3}{2}}\qquad \qquad \qquad \stackrel{reciprocal}{-\cfrac{2}{3}}\qquad \stackrel{negative~reciprocal}{\cfrac{2}{3}}}](https://tex.z-dn.net/?f=%5Cbf%20%28%5Cstackrel%7Bx_1%7D%7B2%7D~%2C~%5Cstackrel%7By_1%7D%7B7%7D%29%5Cqquad%20%28%5Cstackrel%7Bx_2%7D%7B6%7D~%2C~%5Cstackrel%7By_2%7D%7B1%7D%29%20%5C%5C%5C%5C%5C%5C%20slope%20%3D%20m%5Cimplies%20%5Ccfrac%7B%5Cstackrel%7Brise%7D%7B%20y_2-%20y_1%7D%7D%7B%5Cstackrel%7Brun%7D%7B%20x_2-%20x_1%7D%7D%5Cimplies%20%5Ccfrac%7B1-7%7D%7B6-2%7D%5Cimplies%20%5Ccfrac%7B-6%7D%7B4%7D%5Cimplies%20-%5Ccfrac%7B3%7D%7B2%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bperpendicular%20lines%20have%20%5Cunderline%7Bnegative%20reciprocal%7D%20slopes%7D%7D%20%7B%5Cstackrel%7Bslope%7D%7B-%5Ccfrac%7B3%7D%7B2%7D%7D%5Cqquad%20%5Cqquad%20%5Cqquad%20%5Cstackrel%7Breciprocal%7D%7B-%5Ccfrac%7B2%7D%7B3%7D%7D%5Cqquad%20%5Cstackrel%7Bnegative~reciprocal%7D%7B%5Ccfrac%7B2%7D%7B3%7D%7D%7D)
so, we're really looking for the equation of a line whose slope is 2/3 and runs through (5,2)
