Academic Tutoring
- Thread starter TubaRiver
- Start date
I am an English tutor, and spend an average of about 7 hours tutoring Advanced English II/III each week at my school. I would be happy to assist anyone who requires help with grammar/literature, and can provide verification if needed.
I also know quite a bit of world history as well, though I'm not an actual tutor in that subject.
I also know quite a bit of world history as well, though I'm not an actual tutor in that subject.
I am having issues remembering how to solve simple quadratic equations. If anyone has an easy way to remember how to do standard, vertex, and intercept for equations, I would be grateful. I've looked at it millions of times, and can never remember it.
I don't know how to whistle. Pls halp.
(seriously)
I can snap my fingers like a god so it all balances out.
Besides that:
Map Making 101
Fun with Configuration Files
How I Set Up Server? (for nubs)
Basic engineering advice for any aspiring students.
(seriously)
I can snap my fingers like a god so it all balances out.
Besides that:
Map Making 101
Fun with Configuration Files
How I Set Up Server? (for nubs)
Basic engineering advice for any aspiring students.
Sure thing. For a detailed explanation on each click the spoiler. Don't feel like reading everything then just read the end! If you have any more questions don't be afraid to ask!
The standard form of a "Quadratic Equation" is written as: Ax^2 + Bx + C = 0
So, similar to short a polynomial equaled to zero. They can also look weird to trick here you. But here's how to spot them. Ask yourself these questions:
Is it equaled to zero?
Does it have an (x^2) in it?
Does it have a constant (number without x) or at least a coefficient (number with at least one x)
To solve quadratic equations you really just need to remember the quadratic formula and then as i say. "plug in the points."
Because it would look weird if i typed it, this is the quadratic equation!!
When you are presented a quadratic equation just plug in what you have into the formula and solve. Simple as that for STANDARD Quadratics.
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For VERTEX Quadratics:
It is mostly the same thing with a little tweak. You are only looking for 2 values. X and Y.
You are still given the standard quadratic equations to do so. First, find X. (just because it's easier)
x = -b/2a (looks familiar to the quadratic equation doesn't it? Just without the square root)
Now, that you've done that. Find Y
y = c-b^2/4a (This one is a bit trickier, which is why we solved for the easier one first!)
That's all you are doing for vertex quadratics. Lemme do a small example problem for you if you wish.
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Lastly is intercepts. Which are very similar to finding the vertex's. Why? Because you are only looking for 2 values. Nothing more.
x-intercept: You can try standard simplifying; like move over constants and dividing by coefficients to get x by itself, but sometimes that won't work.
Just use the quadratic formula like you would normally. HINT! If the (x^2) is negative. THERE WILL BE NO X-INTERCEPTS! You will wind up with imaginary numbers, which cannot be x-intercepts.
For y-intercepts just plug in 0 into all the x's and solve normally.
4(0)^2 + 3(0) - 5 = 0
well..... everything cancels except the -5 right? Because anything times 0 is 0. So your y-intercept is -5
So, similar to short a polynomial equaled to zero. They can also look weird to trick here you. But here's how to spot them. Ask yourself these questions:
Is it equaled to zero?
Does it have an (x^2) in it?
Does it have a constant (number without x) or at least a coefficient (number with at least one x)
To solve quadratic equations you really just need to remember the quadratic formula and then as i say. "plug in the points."
Because it would look weird if i typed it, this is the quadratic equation!!
When you are presented a quadratic equation just plug in what you have into the formula and solve. Simple as that for STANDARD Quadratics.
_________________________________________________________________
For VERTEX Quadratics:
It is mostly the same thing with a little tweak. You are only looking for 2 values. X and Y.
You are still given the standard quadratic equations to do so. First, find X. (just because it's easier)
x = -b/2a (looks familiar to the quadratic equation doesn't it? Just without the square root)
Now, that you've done that. Find Y
y = c-b^2/4a (This one is a bit trickier, which is why we solved for the easier one first!)
That's all you are doing for vertex quadratics. Lemme do a small example problem for you if you wish.
Find the vertex of the following quadratic:
x^2 + 3x = 0 (remember we are only looking for 2 values)
Our A = 1 because there is nothing right next to it. Our B = 3, and we don't have a C do we? so C = 0
Now for X
x = -b/2a
x = -3/2(1)
2 times 1 = 2
x = -3/2 DONE
Now for Y
y = c-b^2/4a
y = 0-(3)^2/4(1)
3 squared = 9
4 times 1 = 4
0 minus 9 = -9
y = -9/4 DONE
FINAL ANSWER
(-3/2, -9/4)
x^2 + 3x = 0 (remember we are only looking for 2 values)
Our A = 1 because there is nothing right next to it. Our B = 3, and we don't have a C do we? so C = 0
Now for X
x = -b/2a
x = -3/2(1)
2 times 1 = 2
x = -3/2 DONE
Now for Y
y = c-b^2/4a
y = 0-(3)^2/4(1)
3 squared = 9
4 times 1 = 4
0 minus 9 = -9
y = -9/4 DONE
FINAL ANSWER
(-3/2, -9/4)
Lastly is intercepts. Which are very similar to finding the vertex's. Why? Because you are only looking for 2 values. Nothing more.
x-intercept: You can try standard simplifying; like move over constants and dividing by coefficients to get x by itself, but sometimes that won't work.
Just use the quadratic formula like you would normally. HINT! If the (x^2) is negative. THERE WILL BE NO X-INTERCEPTS! You will wind up with imaginary numbers, which cannot be x-intercepts.
For y-intercepts just plug in 0 into all the x's and solve normally.
4(0)^2 + 3(0) - 5 = 0
well..... everything cancels except the -5 right? Because anything times 0 is 0. So your y-intercept is -5
For standard quadratics just use the following formula, and that's it!
For vertex's you are only finding 2 values. X and Y use the formulas to find them, that's it!
x = -b/2a
y = c-b^2/4a
For Intercepts same thing as vertex's ONLY FINDING 2 values. X and Y.
y-intercept: plug in 0 into all x's in the equation and solve. HINT! usually you will wind up just with the constant C as your answer
x-intercept: standard simplifying for X, if it doesn't work, use quadratic formula. HINT! If you have a (-x^2) there will be NO x-intercepts. Because either way, you will end up square rooting a negative number..... they lead to imaginary land! lol
Holy fuck. I remember this:
Our teacher had a song/tune to go along with it that was "Pop goes the weasel." o. o
"X, equals, ne-ga-tive Bee. Plus or minus Bee-Squared. Sub-trac-ting Four-Ayy-See. OV-ER Two-Aa-ayy."
Our teacher had a song/tune to go along with it that was "Pop goes the weasel." o. o
"X, equals, ne-ga-tive Bee. Plus or minus Bee-Squared. Sub-trac-ting Four-Ayy-See. OV-ER Two-Aa-ayy."
C+? Don't think there is such a thing. If so let me know! haha
I know there's C, C#, C+-, and C++
C is just your average run of the mil, general programming language. Fairly easy to learn, and i like to compare it to python! Even though python is a dynamic language and its own compiler all in one. So, no clue why i do that!
C# is really just a Microsoft language. Personally, i don't enjoy C# it's fairly large in usefulness but, its slow, and overall quite weak.
C+- is..... basically a joke. lol Not kidding, it came around to make fun of C and C++.
Finally, we get to C++. The object-oriented language. You could say it's the advanced form of C. Like Java, it is one of the most frequently used languages for many applications. Although, people moan and groan about it because it's "apparently" more difficult to understand compared to Java.
If you really wish, i'd be glad to teach you a bit of the basics for it!
Whenever works for you man, it's not hard to whistle. All I can do is learn you how to do it, improving it will be up to you afterwards. Noone taught me how to whistle, yet I can c:Unfortunately I won't be able to make it (work schedule) but I appreciate the offer.
If you could teach me some coding. That would be awesome! If C++ is the most usefull one then please teach me thatC+? Don't think there is such a thing. If so let me know! haha
I know there's C, C#, C+-, and C++
C is just your average run of the mil, general programming language. Fairly easy to learn, and i like to compare it to python! Even though python is a dynamic language and its own compiler all in one. So, no clue why i do that!
C# is really just a Microsoft language. Personally, i don't enjoy C# it's fairly large in usefulness but, its slow, and overall quite weak.
C+- is..... basically a joke. lol Not kidding, it came around to make fun of C and C++.
Finally, we get to C++. The object-oriented language. You could say it's the advanced form of C. Like Java, it is one of the most frequently used languages for many applications. Although, people moan and groan about it because it's "apparently" more difficult to understand compared to Java.
If you really wish, i'd be glad to teach you a bit of the basics for it!
I need help with basic addition!>>
Joking, I do however, need some help with Algebra I if you have the chance, I have a C+ And anything C is un-acceptable for my capability! But do you think maybe you could help me in a PM?
A PM will be mighty fine :) I'm not the most active of folks right now (On Holiday) but I'll put in extra effort to help.I need help with basic addition!>>
Joking, I do however, need some help with Algebra I if you have the chance, I have a C+ And anything C is un-acceptable for my capability! But do you think maybe you could help me in a PM?
Shoot your questions away!
Perhaps touch on how Holden feels about his brother, whom he considers to be a great writer, working in an industry that he considers "phony" and how this affects him personally. I would also go into detail on how Holden might feel that the movie industry is giving kids like Phoebe false hopes and a skewed vision of reality.
PM me a rough draft or an outline and I can give you a few pointers as well, whenever you get to that point.