I need help with these Algebra 2 questions?

Question

1. The river boat, Alert, take six hours to travel 120 km upstream but only four hours to return. Find the speed of the current and the speed of the riverboat in still water.

Answer: 5km/h Current , 25 km/h riverboat
( I can not get the answer from the question )
Can some1 teach me how to find the answers step by step?


My other question is this.. ( i don't think we even learned this yet )

2. f(x) = x(square) - 1 and g(x) = x +1 , find f(g(x)) and g(f(x))?
( I don't get what this is .. but the answers are )

Answers: f(g(x)) = x(square) + 2x , g(f(x)) = x(square)




THANK YOU VERY MUCH !!

Answer 1

When a boat is rowed in water , its speed will be much higher when it is being rowed with the current , than when it is being rowed against the current. This is like when you ride on a bicycle with the wind behind you , compared to when the wind is blowing against you.

So let us take the speed of the boat in still water as x km. / hr.

Let the speed of the current be y km. / hr.

When the boat is travelling in the same direction as the current , its speed will be x + y. When it is going in the other direction , its speed will be x - y.

Since the distance is given as 120 km. , the time taken in one direction will be 120/(x + y) , whereas the time taken in the other direction will be 120/(x - y).

We have 120/(x + y) = 4 , while 120/(x - y) = 6 [ This is because (x + y) > (x - y). So when the denominator is bigger , the value of the fraction will be smaller ]

Thus , we have two equations :

120 = 4x + 4y ; 120 = 6x - 6y. Solving these two equations , we get x = 25 and y = 5.

2. The problem gives the expressions for two functions :

a) f(x) = x^2 - 1
b) g(x) = x + 1

If we want f( g(x) ) , what it means is that we substitute any occurrence of x in the original equation by the value of g(x) , which is (x + 1).

As an example , suppose we wanted to find out f(7). We would replace x by 7 to get f(7) = 7^2 - 1 = 48.

So when we substitute x by (x + 1) , we get :

f( g(x) ) = (x + 1)^2 - 1 = x^2 + 2x + 1 - 1 = x^2 + 2x.

Similarly , if we want g( f(x) ) , we substitute every occurrence of x in g(x) by the value of f(x) which is (x^2 - 1)

So g( f(x) ) = (x^2 - 1) + 1 = x^2.

Answer 2

1.

A) we move 120 km in 6 hours going against a current of unknown speed, so let's call that "X"

our upstream speed is: 20 km/h

B) Downstream we move 120km in 4 hours with the current.

our downstream speed is: 30 km/h

To find the current.... Let's write equations for both scenarios a, and b

A) 20 + x = 120
B) 30 - x = 120

since we know the distance travled in A) and B) is equal:

we can relate: 20 + x = 30 - x

simplify: x = 10 - x ; 2x = 10 ; x = 5

the speed of the current is then 5 km/hour

You then know what the riverboat is going 25 km/h



2. f(x) = x² - 1 ; g(x) = x+1

Let's find: f( g(x) ) first ...

[1] f ( g(x) ) = f ( x+1) = (x+1)² - 1 = x² + 2x

Now find: g( f(x) )

[2] g ( f(x) ) = g( x² - 1) = ( x² - 1 ) + 1 = x²

The trick: basically replace the input with what you are evaluating... so if you know f ( x ) = 5x , that tells you that the function F takes your INPUT (x) and multiplies it by 5.

Now what about for something else... what does F do to "a+b+c" ?

F ( a + b + c ) = 5 * (a + b + c ) = 5a + 5b + 5c...

So you see writing this all out mathematically you visualize how these things are replaced. It's not complicated...

Take your time and go through all the replacements.

Answer 3

Alert's speed upstream is 120 km / 6 h = 20 km/h; downstream, 120 km / 4h = 30 km/h. Since the speed of the current, v, is speeding up Alert at downstream, slowing up Alert at upstream, and supposing that Alert, by itself, would maintain a steady speed,

30 - v = 20 + v (each side is Alert's normal speed)
10 = 2v
v = 5 km/h

So, Alert, by itself, runs at 20 + 5 = 25 km/h.

To the other question.

f(x) = x^2 - 1
g(x) = x + 1

To combine the functions, just use one as the argument for the other:

f(g(x)) =
(g(x))^2 - 1 =
(x + 1)^2 - 1 =
x^2 + 2x + 1 - 1 =
x^2 + 2x

g(f(x)) = f(x) + 1 = x^2 - 1 + 1 = x^2

Answer 4

The first one is simple:

Ok, we have a riverboat going upstream 120km and it takes 6 hours, well:

120km/6hrs = 20km/h

Now, the riverboat is going downstream, and let's just say it's going the same distance, but this time it only takes four hours, now we have:

120km/4hrs = 30km/h

To find the speed of the boat in still water, we take the average of the upstream and downstream speeds:

30 + 20 = 50

50/2 = 25km/h for the boat

Now we take the 25km/h and subtract it from the downstream speed to find the speed of the current:

30km/h - 25km/h = 5km/h for the current

And there you go.

Answer 5

1)
Let the speed of the river be R and the speed of the current C
Then:
R - C = 120/6 = 20
R + C = 120/4 = 30
Add these two equations together and you get

2R = 50
so R=25mph which makes C = 5mph

(but you're on water - should be using knots :-) )

2)
f(g(x) = (x+1)^2 - 1 = x^2 + 2x = x(x+2)
g(f(x)) = (x^2+1) - 1 = x^2

Answer 6

ok. i don't quite understand the first one.

But the second one is this:

(* ^ means squared*)

f(x)=x^2 - 1
g(x)= x+1

Right? Well, for f(g(x)) (pronounced f of g of x) you take your g(x) and put it in for the x of the f(x). So.....

f(x) = x^2 - 1
(X+1) ^2 - 1
(X^2 + 2X + 1) - 1

so you're left with x squared + 2x + 1 - 1.
1-1 = 0
so ur answer for f(g(x)) is X^2 +2x

______________________________________________________________________________________________

for gf(x)) you do the same thing except you plug in the f(x) equation for all the x's in the g(X).

g(f(x))

g(x) = X+1
f(x) = X^2 -1

so g(f(x)) = (X^2 - 1) +1

well that leaves x sqaured plus one minus one.

x^2 +1-1 which equals just X^2 ---> x-squared.

it's not so bad once you do it a few times. if you need any help in algebra just e-mail me. I'm a college algebra tutor. :D

now about number 1....... that's physics crap. i'll see what i can do.