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Anyone good with Algebra?

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dragonfly411

dragonfly411

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I am in need of some help getting formulas down and learning to figure things. I know some things but the new materials in college algebra aren't sticking and I really think it's because of the teacher. He is very fast paced, generally just shows us a basic formula but doesn't work a problem all the way through as an example, gives no review work, and relies heavily on the calculator. When we ask questions he gets impatient. I'm a bit lost and have a test Friday and I just am not catching everything with quadratics, and figuring rate of time/ height after so many seconds etc. Parabola type stuff. Anyone have any tips, tricks, study sites, anything??
 
Dragonfly, does your college/university have people who tutor on campus? When I was in college they would tutor for free in our library. I would utilize that service if they offer it. The tutors can give you tips and tricks based on what you know or don't know already. good luck Friday!
 
Ditto Skippy's Tutor suggestion. Most likely that tutor will be familiar with your particular teacher's strengths and weaknesses and will have developed a coping mechanism to share with you. Also try hotmath.com for help.

Good luck!
 
I tutor math - I love this stuff. If you want, feel free to post questions and I will try to help you out.
 
I thought an algebra was something worn by a mermaid.
 
Thank you guys for the website suggestions.

Our campus does offer tutors but the tutoring hours are during my work hours, so I'm unable to attend. :(sad

Lady Disdain - Can you explain linear regressions for me? Also, if I get a problem for the finding a height at x time frame (A ball is thrown from the ground, what is the max height, was is the height at n seconds type) could you walk me through that? I know there is more too... I'm going to try to review on our math lab online but it just hasn't been sticking.
 
I've used and liked the algebra tutorials from West Texas A&M. They are layed out in easy to select sections.

Link
 
Hoping Lady Disdain can help you. I stopped doing that type of math 4 years ago and I did them in Swedish so I would probably just confuse you. One tip I have is do a drawing with the projectile, origin hight, trajectory, speed, etc put down everything you know, what you need to know and the different formulas. That usually helps me get what I need to do.

If you have the opportunity do some fysical experiments, it helps you understand it much better and its fun. I remeber one where you put a slide for a marble on top of a table and then by measuring and calculating you find the right spot to put a tiny cup (just a bit bigger than the marble) on the floor. You then send the marble down the slide and cross your tumbs it ends up in the cup, otherwise redo the calculations.

Yikes I sound a bit like a nerd! But really those calculationes are quite fun,I used to like the ones involving bullets. If you think these are boring I hated calculating stuff like what is the temparature of a spoon that you put in a mug of boiling water. Basically we all have likes and dislikes but once you get the hang of it most of it is really fun. Kinda starting to miss math.
 
Linear regressions is a statistics topic. Within the limits of an algebra class, it probably means that you are supposed to "find" a "best fit" line to a set of data. And... maybe they don't really tell you how to find it, other than "guess" or "eyeball it." This is actually a pretty easy to understand presentation, which I did not check for accuracy http://www.google.com/url?sa=t&sour...mjLuJ9MIg&sig2=_Jkj8ehsqvZwRX4GjfGSYA&cad=rja

For the acceleration question, can you post a specific problem and answer that you don't understand?
 
I'll grab a specific problem for the height one and acceleration one. Also dealing with the fact that I can't find my TI83 :((
 
You don't need a calculator when you have your computer... :p

Excel?
 
Stone-cold11|1296749982|2841876 said:
You don't need a calculator when you have your computer... :p

Excel?
Click to expand...

The standard Windows calculator has an option in it to change what it does. I've got mine set to scientific and it does many of the things you need. (not all... great graphing sites and others online if you do a search)
 
We're required to have the TI83 for the tests which are in class, and for the exams also in class. Not allowed to use any computers, phones, or other gadgets.
 
Does your school rent them? I think it's something like $35 for the quarter here. (not exactly cheap, but beats having to buy another $100+ calculator and then finding the other one)
 
Too = I'll check into that thanks!
 
Whoa! I just saw this thread. My daughter is taking Algebra II in college this year. (She took it in high school, but managed to test into it when she did the placement test for college anyway.) I wonder if she is too lazy to use this thread?

dragonfly, you are an inspiration!!!

PS-I used the free tutor available when I took Physics during my freshman year in college. Chemistry (in high school) had been my favorite subject. Little did I know that Physics would be anathema to me!

Big hugs,
Deb
:read:

Diamonds Are A Girl's Best Friend
 
dragonfly411|1296672386|2840844 said:
Also, if I get a problem for the finding a height at x time frame (A ball is thrown from the ground, what is the max height, was is the height at n seconds type) could you walk me through that? I know there is more too... I'm going to try to review on our math lab online but it just hasn't been sticking.
Click to expand...

You probably begin with some formula for height in terms of time, h(t).

In order to find the height at n seconds, plug n in wherever you see the t, you are finding h(t=n).

If a ball is thrown from the ground, that is, at t=0, h=0, and you want to find the maximum height, you must first find the time that the ball hits the ground on the way down. Set h(t)=0, and you may get two answers; you are interested in the positive, nonzero answer, which I will call t'. Since the ball is at its maximum height at half of the time of the trajectory (it spends half the time going up, the other half going down) you are interested in h(t'/2).

HTH
 
Julie - Thanks, That makes sense.

I promise I'm going to get an actual question to post, I've been at work so no book with me.
 
OK Here is the example.

A baseball is hit straight up with an initial velocity of v= 80 ft per second and leaves teh bat with an initial height of 3 ft.
a) write a formula that models the height of the baseball after t seconds
b) how high is the baseball after 2 seconds
c) find the maximum height of the baseball, support answer graphically (I'm assuming we graph on the calculator then draw it out on a diagram IF the graph is required on the test. I really wish they'd just let us know how to manually figure out the graph)
 
do you have the answers?
 
dragonfly411|1296785243|2842425 said:
OK Here is the example.

A baseball is hit straight up with an initial velocity of v= 80 ft per second and leaves teh bat with an initial height of 3 ft.
a) write a formula that models the height of the baseball after t seconds
b) how high is the baseball after 2 seconds
c) find the maximum height of the baseball, support answer graphically (I'm assuming we graph on the calculator then draw it out on a diagram IF the graph is required on the test. I really wish they'd just let us know how to manually figure out the graph)
Click to expand...

a) h(t)= initial height + initial upward/downward velocity*t + 1/2*acceleration*t^2
h(t)= 3 ft + 80 ft/s *t + .5 * (-32.17 ft/s^2) *t^2

b) h(t=s) = 3 + 80*2 -.5*32.17*4= 98.66 ft

c) h(t)=0=3 + 80t -16.09t^2. Use the quadratic formula. t = {-80 +/ sqrt [80^2-4*(-16.09)(3)]} / (-32.17) = -.037, 5.011. The first answer is negative because you started with an initial height of 3. The second answer, 5.01, is when the ball hits the ground coming down. Earlier, I said that you're supposed to take this and divide by two, but that only applies if the ball started at h=0... I said that it must spend half the time going up, and half the time going down. However, since the ball started at h=3, it got a head start on the going up part, and it had to spend a little extra time going down. You don't want to count this extra time it has to go the 3 feet down, so t'=5.011-.037 = 4.974 s
t'/2= 2.487 s

OR

velocity = initial velocity + acceleration * t = 80 ft/s - 32.17 ft/s^2. Set equal to zero and solve for t.
80 - 32.17t = 0
80 = 32.17t
t = 2.487 s

h(2.487)= 102.47 ft

The way you normally manually graph a parabola is to graph the vertex and the zeros, or the vertex and some points around the vertex.
If I were asked to support my answer graphically, I would graph height versus time, and label the vertex.
 
simple kinematic equations.

u = initial velocity
v = final velocity
a = acceleration (-negative of gravitational acceleration)
t = time
s = displacement (height in this case)
h = initial displacement

Simple to derive using integration.

v = u + a * t
s = h + u * t + 1/2 * a * t ^2

b) Sub in the values and t = 2s into the displacement equation.

c) Maximum height reach is when v = 0, so use this condition on v = u + a t to find the time taken to reach v = 0

v = 0 = u + a t
-u = a t
t = -u / a

And use the t and sub into the displacement equation. to get the maximum height reached.

Manually graph. This is a quadratic equation, just plot the values for different time on a scale graph and then draw a smooth line through all the points and read off the maximum from there.
 
The basic equation relating height of a projectile as a function of time is:
h(t)=h(initial)+v(initial)*t-1/2*a*t^2 where a=acceleration due to gravity

Here on earth, where a is approx. 32 ft/sec^2, the equation becomes:
h(t)=h(initial)+v(initial)*t-16t^2

So, for your problem, where initial height is 3 and initial velocity is 80, your equation is:
h(t)=3+80t-16t^2
h(t) is just function notation for y. In your graphing calculator, you can enter it as y=3+80t-16t^2

The shape of this curve is an upside down parabola. In order to graph it manually, you need to find "critical points". For a parabola, these points are the x-intercepts, the y-intercepts and the vertex. Y-intercept is the easiest; plug in 0 for x and solve for y; you get (0,3). To find the x intercepts, plug in 0 for y and solve for x: 0=-16t^2+80t+3. You would need to use the quadratic formula for this as the equation is not factorable. To find the vertex, first find the axis of symmetry which is given by the formula x=-b/2a, or in this case -80/(2*-16) which comes to x=2.5. The vertical line x=2.5 is the line that this upside down parabola is symmetrical about. That means it hits its peak at x=2.5 seconds. To find out how high it went in 2.5 seconds, just plug in 2.5 for t to get the height.

Hope this helps!

edited to add: LOL, we were all 3 posting at about the same time! What a bunch of geeks we are!
 
Maria D|1296789177|2842494 said:
The basic equation relating height of a projectile as a function of time is:
h(t)=h(initial)+v(initial)*t-1/2*a*t^2 where a=acceleration due to gravity

Here on earth, where a is approx. 32 ft/sec^2, the equation becomes:
h(t)=h(initial)+v(initial)*t-16t^2

So, for your problem, where initial height is 3 and initial velocity is 80, your equation is:
h(t)=3+80t-16t^2
h(t) is just function notation for y. In your graphing calculator, you can enter it as y=3+80t-16t^2

The shape of this curve is an upside down parabola. In order to graph it manually, you need to find "critical points". For a parabola, these points are the x-intercepts, the y-intercepts and the vertex. Y-intercept is the easiest; plug in 0 for x and solve for y; you get (0,3). To find the x intercepts, plug in 0 for y and solve for x: 0=-16t^2+80t+3. You would need to use the quadratic formula for this as the equation is not factorable. To find the vertex, first find the axis of symmetry which is given by the formula x=-b/2a, or in this case -80/(2*-16) which comes to x=2.5. The vertical line x=2.5 is the line that this upside down parabola is symmetrical about. That means it hits its peak at x=2.5 seconds. To find out how high it went in 2.5 seconds, just plug in 2.5 for t to get the height.

Hope this helps!

edited to add: LOL, we were all 3 posting at about the same time! What a bunch of geeks we are!
Click to expand...


Ok so to continue with this, (since I bombed my test over a question like this) what if you are given a problem as follows:

A grasshoper is on a blade 6 in high, and jumps. He lands 10.5 in away from the blade. I was given the equation :

H= -.2X^2 - 1.5X + 6 with x equal to distance from the blade and h equal to height. And the question asks how far away from the blade was the grasshoper when he is 4.75 inches high.
 
And can someone explain the 32.17 or how to derive the acceleration due to gravity?
 
dragonfly411 said:
And can someone explain the 32.17 or how to derive the acceleration due to gravity?
Click to expand...

That is a constant.

For the grasshopper question, just substitute the h = 4.75 into the equation and solve for x.
H= -.2X^2 - 1.5X + 6

4.75 = -.2x^2 - 1.5x + 6
.2x^2 + 1.5x - 1.25 = 0

using x = (-b +- SQRT(b^2 - 4ac))/2a

where a = 0.2, b = 1.5, c = -1.25
 
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