# Math Help - Application Problems

1. ## Application Problems

1.) The length-weight relationship for the pacific Halibut can be approximated by the function L(w)=.46*cube-root(w), where w is the weight in kilograms and L is the length in meters. If the halibut's weight is 230 kilograms, what is its length?

Okay seems simple enough and I set it up like this:

L(w)=.46*cube-root(260)

Well, if you cube both sides you end up with 25.30736 but with a cube on the other side. It puts you back in the same spot. Any ideas?

2.) A truck on the highway is traveling at 55 mph. Three hours later, a car gets on the same highway at the same place as the truck and begins traveling in the same direction at 75 mph. After what distance will the car overtake the truck?

I don't know how to set this equation up.

3.) It takes a boy 90 minutes to mow the lawn, but his sister can mow it in 60 minutes. How long would it take them to mow the lawn if they worked together using two lawn mowers?

Here is how I set this up:

1/90+1/60=1/t and solved for t, which gave me 25.7 minutes. Does that seem right?

2. Originally Posted by ZE2001
1.) The length-weight relationship for the pacific Halibut can be approximated by the function L(w)=.46*cube-root(w), where w is the weight in kilograms and L is the length in meters. If the halibut's weight is 230 kilograms, what is its length?

Okay seems simple enough and I set it up like this:

L(w)=.46*cube-root(260)

Well, if you cube both sides you end up with 25.30736 but with a cube on the other side. It puts you back in the same spot. Any ideas?
w = 230 or 260 ??? Let us take w = 230 kg

$L(w) = 46.\sqrt[3]{230}$

take cube-root of 230 using calculator,

$L(w)= 46 \times 6.1269$

$L(w) = 281.8$

Did you get it now???

3. Originally Posted by Shyam
w = 230 or 260 ??? Let us take w = 230 kg

$L(w) = 46.\sqrt[3]{230}$

take cube-root of 230 using calculator,

$L(w)= 46 \times 6.1269$

$L(w) = 281.8$

Did you get it now???
Yes I understand that, but I don't think I have a cube root button on my TI-30X IIS.

4. Originally Posted by ZE2001
2.) A truck on the highway is traveling at 55 mph. Three hours later, a car gets on the same highway at the same place as the truck and begins traveling in the same direction at 75 mph. After what distance will the car overtake the truck?

I don't know how to set this equation up.
The distance travelled by truck in 3 hours = speed . time = (55) (3) = 165 miles.

Let the car overtakes the truck after t hours.

Distance by car in time t = 165 + distance by truck in time t

75t = 165 + 55t

Now, solve this eqn for t,
we get t = 8.25 hours = 8 h 15 min

Now, distance travelled by car = (75)(8.25)= 618.75 miles.

The car overtakes truck after 618.75 miles from starting point.

Did you get it ???

5. Originally Posted by Shyam
The distance travelled by truck in 3 hours = speed . time = (55) (3) = 165 miles.

Let the car overtakes the truck after t hours.

Distance by car in time t = 165 + distance by truck in time t

75t = 165 + 55t

Now, solve this eqn for t,
we get t = 8.25 hours = 8 h 15 min

Did you get it ???
Yes that makes sense. I really struggle with setting these up so thank you!

6. Originally Posted by ZE2001
Yes I understand that, but I don't think I have a cube root button on my TI-30X IIS.
use the calculator which is on your computer.

go to start--> all programs -->accessories-->calculator.

once you open calculator, go to, view-->scientific.

7. Originally Posted by ZE2001
3.) It takes a boy 90 minutes to mow the lawn, but his sister can mow it in 60 minutes. How long would it take them to mow the lawn if they worked together using two lawn mowers?

Here is how I set this up:

1/90+1/60=1/t and solved for t, which gave me 25.7 minutes. Does that seem right?
your eqn is right but it will give t = 36 minutes.

see,

$\frac{1}{90} + \frac{1}{60} = \frac{1}{t}$

$\frac{2}{180} + \frac{3}{180} = \frac{1}{t}$

$\frac{5}{180} = \frac{1}{t}$

$t = \frac{180}{5}$

$t = 36\;\;min$

it will take 36 min if they work together.