Word Problems

[nycmath]

See attachment.

[Eigenvalue]

1. First season: 1382
Second season:1256
Third season=1256-262

Total number of tickets=First season+second season+third season
=1382+1256+1256-262
=1382+1256+994
=1382+2260
=3642

[nycmath]

1. First season: 1382
Second season:1256
Third season=1256-262

Total number of tickets=First season+second season+third season
=1382+1256+1256-262
=1382+1256+994
=1382+2260
=3642

I'm having trouble with question 2. What is the setup?

[Eigenvalue]

2. Use the formula speed=distance/time
John's speed=5mph
Lucy's speed: 3.5mph

Let the time since 10:00 be t
John's time: t-0.5
Lucy's time: t
Note: for Lucy's time, use "t-0.5" rather than "t-30," as the unit for speed is meters per hour

The distance is the same. Distance=speed*time
5(t-0.5)=3.5t
5t-2.5=3.5t
1.5t=2.5
t=[tex]\frac{5}{3}[/tex]hours=100 minutes

[nycmath]

2. Use the formula speed=distance/time
John's speed=5mph
Lucy's speed: 3.5mph

Let the time since 10:00 be t
John's time: t-0.5
Lucy's time: t
Note: for Lucy's time, use "t-0.5" rather than "t-30," as the unit for speed is meters per hour

The distance is the same. Distance=speed*time
5(t-0.5)=3.5t
5t-2.5=3.5t
1.5t=2.5
t=[tex]\frac{5}{3}[/tex]hours=100 minutes

Why did you manipulate the normal formula for distance
d = (speed)(time)? Why did you solve the formula for speed? Is there a particular reason why you needed to do that?

[Eigenvalue]

2. Use the formula speed=distance/time
John's speed=5mph
Lucy's speed: 3.5mph

Let the time since 10:00 be t
John's time: t-0.5
Lucy's time: t
Note: for Lucy's time, use "t-0.5" rather than "t-30," as the unit for speed is meters per hour

The distance is the same. Distance=speed*time
5(t-0.5)=3.5t
5t-2.5=3.5t
1.5t=2.5
t=[tex]\frac{5}{3}[/tex]hours=100 minutes

Why did you manipulate the normal formula for distance
d = (speed)(time)? Why did you solve the formula for speed? Is there a particular reason why you needed to do that?

s=[tex]\frac{d}{t}[/tex] and d=st are mathematically equivalent; I just used the form s=[tex]\frac{d}{t}[/tex]

[nycmath]

2. Use the formula speed=distance/time
John's speed=5mph
Lucy's speed: 3.5mph

Let the time since 10:00 be t
John's time: t-0.5
Lucy's time: t
Note: for Lucy's time, use "t-0.5" rather than "t-30," as the unit for speed is meters per hour

The distance is the same. Distance=speed*time
5(t-0.5)=3.5t
5t-2.5=3.5t
1.5t=2.5
t=[tex]\frac{5}{3}[/tex]hours=100 minutes

Why did you manipulate the normal formula for distance
d = (speed)(time)? Why did you solve the formula for speed? Is there a particular reason why you needed to do that?

s=[tex]\frac{d}{t}[/tex] and d=st are mathematically equivalent; I just used the form s=[tex]\frac{d}{t}[/tex]

I understand. Thanks.