On a windy day william found that he could travel 10 mi downstream and then 2 mi back
upstream at top speed in a total of 16 min. what was the top speed of william's boat if the rate
of the current was 30 mph? (let x represent the rate of the boat in still water.)

The top speed of William's boat was 45 mph

Step-by-step explanation:

Let

x -----> represent the rate of the boat in still water in mph

we know that

The speed or rate is equal to divide the distance by the time

speed=distance /time

time=distance/speed

Downstream

speed=(30+x) mph

distance=10 mi

time1=10/(30+x)

Upstream

speed=(x-30) mph

distance=2 mi

time2=2/(x-30)

The sum of the time downstream plus the time upstream must be equal to 16 minutes

Convert minutes to hours

Multiply by (x+30)(x-30) both sides

Multiply by 60 both sides

Divide by 16 both sides

The solution is x=45\ mph

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.

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