# Hot-Air Balloon

### By: Abishek Balakrishnan and Zubair Siddiqui (Period 6)

## Question 1

## Question 2

## Question 3

*before*the balloon was spotted, and this can be rewritten as -5 minutes after the balloon was spotted. Then we do the work:

1.) f(x)= 800-20(-5)

2.) f(x)= 800- (-100), which is the same as f(x) = 800+100

3.) f(x)= 900

This gives us the answer, **900 feet.**

## Question 4

1). 20= 800-20x

2.) Subtract 800 from both sides to get, -780=-20x

3.) Now divide both sides by -20 (to isolate the x), and get 39= x

4.) Rewrite 39=x, and get x= 39 (number=x can always be rewritten as x=number)

This leads us to the answer, **39 minutes.**

## Question 5a

**f(x) = 1200-20x.**

## Question 5b

However, this is only the first part of the question. Now we have to see how much longer it will take for the second balloon to land than the first one. Landing tells us f(x) will be 0 (see Question 1 for why). We just have to plug this into both functions, and find the difference. For the first function:

1.) 0=800-20x

2.) Subtract 800 from both sides, and get -800=-20x

3.) Divide both sides by -20 (to isolate x), and get *40=x*

4.) **40 minutes**

For the second function:

1.) 0=1200-20x

2.) Subtract 1200 from both sides, and get -1200=-20x

3.) Divide both sides by -20, and get *60=x*

4.) **60** **minutes**

Now we subtract to see how much longer it took for the second balloon to land.

1.) 60-40=20. It took **20 more minutes** for the second balloon to land!

## Question 6

**f(x) = 800-30x.**Now, using what we did in Question 5b, we must find how much longer it'll take for the third balloon to land than the first one. We already have the answer to the first function (40 minutes to land), so we only need to calculate how long it takes to land for the third one.

1.) 0=800-30x

2.) Subtract 800 from both sides to get -800=-30x

3.) Divide both sides by -30, and get *26 2/3=x*

4.) **26 minutes and 40 seconds **(same thing as mixed fraction above)

Next, we do 26 2/3 -40=-13 1/3** . **It took

**13 minutes and 20 seconds fewer**minutes for the third balloon to land than the first one (negative more minutes is the same as positive less minutes)!

## Question 7

**f(x) = 30x.**Next, the question is asking for when the first and fourth balloons will be at the same altitude. To show this, we make the two functions equal each other. Since both functions equal f(x), we can leave this out, and just write the rest. This would be

**800-20x=30x.**This is how we solve this equation:

1.) 800-20x=30x

2.) Subtract -20x from both sides (same as adding 20x to both sides) to get the x's all on one side. This will get you 800=50x

3.) Divide both sides by 50, and get 16=x

This means that both balloons would have the same altitude after** 16 minutes **of the first balloon being spotted and the fourth one being launched. To find the altitude, we would just plug 16 in place of x. It could be either **y=30(16), or y=800-20(16), which equals 480 feet. **(y is the altitude). The equations would equal the same thing because they intersect after 16 minutes.

## Question 8

In this question, we are trying to calculate how high the third balloon should be if we want it to land at the same time as the first balloon. We can use a simple equation to help us find this out. **y=mx+b. **This equation is called the slope intercept equation. We have y, which is 0, because when they land the are both on the ground and not up in the air. We also have m and x. M is the feet per minute it falls, which is -30, and x is how long it takes to land, which is 40. Now we just plug the numbers in and we get out equation.

1.) We just substitute the numbers. 0=-30(40)+b

2.) We now solve the equation. -30 times 40 equals -1200. 0=-1200+b

3.) We finish the equation by adding 1200 to both side to get 1200=b

4.) We now found our answer. The third balloon would have to be** 1200 feet in the air** for it to land at the same time as the first balloon. Thus, the equation for this line would be **y=-1200+b.**