Any algebra whizzes???

[krcit]

My dd is having trouble with 2 math problems and I'm useless in math.


One leg of a right triangle exceeds the other leg by 4 inches. The hypotenuse is 20 inches. Find the length of the shorter leg of the right triangle.



The second problem is:

A batter hit a baseball at a height of 4 feet with a force that gave the ball an initial velocity of 64 feet per second. The equation h=-16t2 +64t +4 gives the height h of the baseball t seconds after the ball was hit. If the ball was caught at a height of 4 feet, how long after the batter hit the ball was the ball caught?



If anyone could give her a hint as to how to figure these out, I'd appreciate it. I can't even understand the questions

 

[Aliceacc]

I teach algebra

1. One leg of a right triangle exceeds the other leg by 4 inches. The hypotenuse is 20 inches. Find the length of the shorter leg of the right triangle.

Let x = the shorter leg, let x+4 = the longer leg.

Since it's a right triangle, you can use Pythagorean Theorem (a squared + b squared = c squared.

x squared + (x+4) squared = 20 squared.

Remind her to use "FOIL" to square the x+4.


2. A batter hit a baseball at a height of 4 feet with a force that gave the ball an initial velocity of 64 feet per second. The equation h=-16t2 +64t +4 gives the height h of the baseball t seconds after the ball was hit. If the ball was caught at a height of 4 feet, how long after the batter hit the ball was the ball caught?

If the ball was caught at a height of 4 feet, then h=4. So the equation becomes 4=16t2 +64t +4. Have her subtract 4 from both sides, then factor. She'll get two answers. The first is when it starts off at 4 feet, the other is the answer she wants.

 

[PSUfanonBTMR]

1st:

a^2 + (a+4)^2=20^2

formula used: a^2+b^2=c^2

The second is very confusing, I agree

 

[taheenee]

the 1st one I can help with, the second one, not so much. I'll have to think on that one and get back to you. on the 1st one, with a right triangle a^2 + b^2 = c^2, this is a squared + b squared = c squared with c being the hypotenuse. if you express one leg as 'a' and the other leg as 'a+4' you can fill in the formula and solve for a.

 

[taheenee]

1st:

a^2 + a^2 +4=20^2

formula used: a^2+b^2=c^2

The second is very confusing, I agree

actually it would be (a+4)^2 for the 2nd part

 

[TinkPinkPoem]

I teach algebra

1. One leg of a right triangle exceeds the other leg by 4 inches. The hypotenuse is 20 inches. Find the length of the shorter leg of the right triangle.

Let x = the shorter leg, let x+4 = the longer leg.

Since it's a right triangle, you can use Pythagorean Theorem (a squared + b squared = c squared.

x squared + (x+4) squared = 20 squared.

Remind her to use "FOIL" to square the x+4.


2. A batter hit a baseball at a height of 4 feet with a force that gave the ball an initial velocity of 64 feet per second. The equation h=-16t2 +64t +4 gives the height h of the baseball t seconds after the ball was hit. If the ball was caught at a height of 4 feet, how long after the batter hit the ball was the ball caught?

If the ball was caught at a height of 4 feet, then h=4. So the equation becomes 4=16t2 +64t +4. Have her subtract 4 from both sides, then factor. She'll get two answers. The first is when it starts off at 4 feet, the other is the answer she wants.

I totally agree. There always seems to be someone catching things up before me on these math threads.

Second problem is basic physics but then again it could be used in a Maths problem.

 

[BuzznBelle'smom]

Another way to look at the first problem:

One possible side combination of a right triangle is 3,4,5 (5 being the hypotenuse). By multiplying through by 4, you get sides of 12,16, and the given 20. However, her teacher is probably not looking for the problem to be solved this way--but it would help your DD to understand what the correct answer is. She really should go through the quadratic equations to find the answer using algebra.

Off to read the second problem...

 

[EthansMom]

My dd is having trouble with 2 math problems and I'm useless in math.


One leg of a right triangle exceeds the other leg by 4 inches. The hypotenuse is 20 inches. Find the length of the shorter leg of the right triangle.

For a right triangle, a^2 + b^2 = c^2 where a is the length of one leg, b is the length of the other leg, and c is the length of a hypotenuse.
b = a + 4
a^2 + (a+4)^2 = (20)^2
solve for a

A batter hit a baseball at a height of 4 feet with a force that gave the ball an initial velocity of 64 feet per second. The equation h=-16t2 +64t +4 gives the height h of the baseball t seconds after the ball was hit. If the ball was caught at a height of 4 feet, how long after the batter hit the ball was the ball caught?

h = 4 feet
4 = -16t^2 + 64t + 4
solve for t

 

[BuzznBelle'smom]

The second equation isn't too difficult, if you plug in h=4 to solve. Reduce, factor, and you're there.

 

[TinkPinkPoem]

Another way to look at the first problem:

One possible side combination of a right triangle is 3,4,5 (5 being the hypotenuse). By multiplying through by 4, you get sides of 12,16, and the given 20. However, her teacher is probably not looking for the problem to be solved this way--but it would help your DD to understand what the correct answer is. She really should go through the quadratic equations to find the answer using algebra.

Off to read the second problem...

Yup, it could be those numbers, but it's just a hypothesis (added: it is those numbers, but I'm making a point!). You can only prove this by multiplying by 4 and frankly it would stand no ground in math as a solution. In another problem context there could be so many solutions that would fit and a hypothesis could be easily dismissed.

I think she should use the Pythagorean Theorem.

 

[BuzznBelle'smom]

Actually, it MUST be those numbers. For the hypotenuse (longest side) of a right triangle to be 20, with the otehr sides being x and x+4, the only possible answer is 12, 16, 20. Using the Pythagorean theorum. However, as I stated, I don't believe the teacher wanted the problem solved this way--but it's a way to check the answer.

 

[TinkPinkPoem]

Actually, it MUST be those numbers. For the hypotenuse (longest side) of a right triangle to be 20, with the otehr sides being x and x+4, the only possible answer is 12, 16, 20. Using the Pythagorean theorum. However, as I stated, I don't believe the teacher wanted the problem solved this way--but it's a way to check the answer.

I agree!

 

[krcit]

Thanks for all the help so far. she just left for basketball practice but I'll have her look at this when she gets back. she's having trouble b/c they haven't yet gone over this in class. I'm glad I'm not in school these days if this is 8th grade math

 

[phinsphan]

No help for the problem but I have a dd in algebra and I have found http://www.bagatrix.com/products.html software worth every $$$ and also we've used www.tutor.com for help as well. Tutor is pricey but I like it.

 

[Aliceacc]

Since both problems involve quadratic equations, I'm guessing that the teacher wants to see the factoring. Knowing the Pythagorean Triples is a help, but I would be more interested in the solution than the answer.

 

[PSUfanonBTMR]

actually it would be (a+4)^2 for the 2nd part

Oops, right

Hope she didn't look at mine