I’m not 100% sure but I think it’s 4 for the gcf
I fear that there is an error copying your assignment.
65% of the students walked. Are all of the other students on buses? Are there both public buses AND private buses?
Assuming that you need to know BOTH kinds of buses, try this:
65% of the students walked, so since 100% - 65% = 35% then this means that 35% of the students were on buses.
Since we know that there are 360 more walkers than bus riders, then one equation we know is: 65% of S = 360 + 35% of S (let S = total # of students)
.65 S = 360 + .35 S
<u> - .35 S </u> = <u> - .35 S</u> Subtract .35 S from both sides
<u> .30 S </u> = <u> 360</u> Divide both sides by .30 (or .3)
.30 .30
S = 1,200 so we know that this is the total number of students, but that is not what was asked.
They want to know how many are on buses and specifically how many are on public buses, if I read this correctly.
Since the walkers = 65% of 1,2000 and we know of means TIMES, then
.65 (1,200) = 780 walkers
1,200 total students minus 780 walkers = 420 bus riders
Now, if there is not a misprint and we really have to figure out the public bus riders as compared to the private bus riders, then remember the ratio from above in the question: 4 bus: 3 public buses
Now if I read this right, that means that 3/4ths of the bus riders were on public buses
so 3/4 of 420 means 3/4 times 420 = 3 times 105 = 315 public bus riders (which coincidentally leaves 105 private bus riders, but since they are private we don't know much about them. Ha-Ha..... I made a lame joke.)
So your answer is 315 public bus riders
It's difficult to make out what the force and displacement vectors are supposed to be, so I'll generalize.
Let <em>θ</em> be the angle between the force vector <em>F</em> and the displacement vector <em>r</em>. The work <em>W</em> done by <em>F</em> in the direction of <em>r</em> is
<em>W</em> = <em>F</em> • <em>r</em> cos(<em>θ</em>)
The cosine of the angle between the vectors can be obtained from the dot product identity,
<em>a</em> • <em>b</em> = ||<em>a</em>|| ||<em>b</em>|| cos(<em>θ</em>) ==> cos(<em>θ</em>) = (<em>a</em> • <em>b</em>) / (||<em>a</em>|| ||<em>b</em>||)
so that
<em>W</em> = (<em>F</em> • <em>r</em>)² / (||<em>F</em>|| ||<em>r</em>||)
For instance, if <em>F</em> = 3<em>i</em> + <em>j</em> + <em>k</em> and <em>r</em> = 7<em>i</em> - 7<em>j</em> - <em>k</em> (which is my closest guess to the given vectors' components), then the work done by <em>F</em> along <em>r</em> is
<em>W</em> = ((3<em>i</em> + <em>j</em> + <em>k</em>) • (7<em>i</em> - 7<em>j</em> - <em>k</em>))² / (√(3² + 1² + 1²) √(7² + (-7)² + (-1)²))
==> <em>W</em> ≈ 5.12 J
(assuming <em>F</em> and <em>r</em> are measured in Newtons (N) and meters (m), respectively).
Lcm of 60 and n = 420
Stated: n <90 and a multiple of 6
420/60 = 7 thus n must be a multiple of 7
Therefore n =42 or 84
