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Lapatulllka [165]
3 years ago
5

A girls' track team must run 3 miles on the first day of practice and 6 miles every day after that. The boys' team must run 5 mi

les every day of practice. The coach will order new javelins at the end of the day that each girl's total mileage surpasses each boy's. How many total miles will each girl have run by the time the coach orders the new equipment?
Mathematics
2 answers:
Vera_Pavlovna [14]3 years ago
7 0

Answer:

Each girl would have ran a total of 21 miles and boys would have ran a total of 20 miles by the end of the fourth day before the couch order new javelins.

Step-by-step explanation:

Girls track team

First day of practice=3 miles

Days afterwards= 6 miles

Boys track team

Everyday= 5 miles

coach will order new javelins at the end of the day that each girl's total mileage surpasses each boy's.

First day

Each girl total mileage= 3 miles

Each boy total mileage= 5 miles

Second day

Each girl total mileage= 3miles+6miles=9 miles

Each boy total mileage=5 miles+5 miles= 10 miles

Third day

Each girl total mileage= 3+6+6=15 miles

Each boy total mileage=5+5+5=15 miles

Fourth day

Each girl today mileage= 3+6+6+6=21 miles

Each boy total mileage=5+5+5+5= 20 miles

Difference=girls-boys

=21-20= 1 mile

The fourth day is the day each girl's total mileage(21 miles) surpasses each boy's total mileage(20 miles) with a total of 1 mile

Elis [28]3 years ago
3 0

Answer:

The 4th day so 21 miles the girls would have to run

Step-by-step explanation:

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Answer:

a) Applying Pithagoras Theorem

b) Trigonometric relations

c) No we can not

d) No we can not

e) See step-by-step explanation

f) See step-by-step-explanation

Step-by-step explanation:

a) The easiest method for calculating the missing side is applying Pithagoras Theorem

c²  =  a²  +  b²

c²  =  (5)²  +  (12)²      ⇒  c²  = 25 + 144    ⇒ c²  =  169    ⇒  c  = 13

b) The easiest method to find the missing angles, is to calculate the sin∠ and then look for arcsin fuction in tables.

sin ∠α = 5/13    sin ∠α = 0.3846    ⇒      arcsin (0.3846 )   α  = 23⁰

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c) The law of cosines should not be applied to right triangles, in fact for instance in our particular case we have:

Question a)

c²  = a² + b² - 2*a*b*cos90⁰   (law of cosines ) but cos 90⁰ = 0

then c²  = a² + b² which is the expression for theorem of Pithagoras

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c²  = a²  +  b²  - 2*5*13*cos α

169 = 25  +  144 - 2*5*13* cos α

As you can see 169 - 25 - 144 = 0

Then we can not  apply law of cosine in right triangles, we shoul apply trigonometric relations and Pithagoras theorem, and as we saw you can get the expression of Pithagoras theorem from cosine law

c²  =  a²  +  b²  - 2*a*b*cos∠90°    cos ∠ 90°  = 0

Then     c²  =  a²  +  b²

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