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Omg im so pissed got a 55/100 on my trig test.

 

Bummer. Any specific help you looking for?

 

Is this just for math work? because I get math, I like math, but chemistry, Different story :\

But I'm really good at english, literature, analysis, practice, theory, symbolism and such, so if anyone needed help there :]

 

This thread is for all sorts of homework help. Math is just usually the easiest to explain and most confusing for a lot of people. If you have any questions at all though, feel free to post!

 

Pretty much looking for all the answers.

http://img253.imageshack.us/img253/3664/georg2.jpg

 

Here ya go:

There is a sticky in this forum for a reason...but give me a few minutes :S

You want the WHOLE worksheet done? I don't have that much time at the moment, but here's a brief explanation of the first few.

1) Reflect the triangle over the y axis - keep the points on the same "height" but basically draw the mirror image. For example, point C (-6,1) because point C' (6, 1). If you fold the paper horizontally, the two triangles should overlap.

2) Reflect the triangle over the x axis - keep the points the same distance from the y-axis, but just shift them down. For example, point C (-6, 1) because point C' (-6, -1). Comprende? They should overlap if you fold the paper over the x-axis.

3) MOVE the triangle 5 units to the right and 6 units down. For example, point C (-6, 1) because point C' (-1, -5).

4) Which variable (x or y) determines the distance horizontally? The answer is x, therefore if you want to move it 5 units to the left, horizontally, you need to decrease the x value by 5. So, (x, y) --> (x-5, y)

5) Since horizontal distance is affected by the x, vertical distance is affected by the y value. Therefore, to move it right 2 you increase x by 2, and to move it up 4 you increase x by 4. So, (x, y) --> (x+2, y+4)

6) To move it down, all you have to do is decrease the y value, right? So what you do is (x, y) --> (x, y-2).

6.5) This is going to be a horizontal reflection.

7) This is going to be a translation, moved 5 units right and down 6.

8) This is going to be a vertical reflection.

9) Well, if you're given the points and the relationship, you can just find the new points. As we know the rule is (x,y) -> (x-2, y+3), and the original points: A (4, 2), B (-3, -4), C(-5, 5), we can just calculate the new points.
A' (2, 5), B' (-5, -1), C' (-7, 8)

10) Okay, since we know that in a triangle, AB + BC > AC, AB + AC > BC, BC + AC > AB (the triangle inequality theorem; the sum of any two sides of a triangle must be greater than the third side - this makes sense, because let's just imagine a straight line. It can be broken up into a "fake triangle" if you take any point on that line and call it B, while calling the endpoints A and B. AB + BC = AC, but if it were any shorter, the segments would not reach the ends - make sense?)

Anyways, we know that in a possible triangle, 9, or x could be the longest side [given sidelengths 4, 9, and x]. So, if x is the biggest side, the outer bound would be: 4 + 9 > x, 13 > x -> x < 13. Therefore, we know that x cannot be bigger than 13. Now, what if 9 is the longest side? Our inequality would be 4 + x > 9, thus x > 5. So, the final possible inequality would be:
5 < x < 13. It cannot be equal to 5 or 13, because otherwise we would just have a line.

11) Angle F must be 57, because the angles of a triangle add up to 180. So, the angles (smallest to largest) must be 33, 57, 90. Now, there is not enough information to find the specific sidelengths, but to order them largest to smallest, we can use this theorem: the longest side is always opposite the largest angle in a triangle.

So, opposite from angle 33 (leg DF) is the smallest, opposite from angle 57 (leg DE) is the next largest, and finally the side opposite from the right angle (leg EF) is the greatest.

12) Well, we know that side with length 3 is across from the 16 degree angle, the side with length 7 is across from the 41 degree angle, and the side with length 9 is across from the 126 angle. I don't really want to draw it, but it's not really that difficult. I might attach a drawing that looks something similar to it later.

13) Can't tell. Picture this: why can't you draw AD or DC in such a way that they intersect line AB/BC much closer to point B? The truth is, you can't - there is no restriction on the length of AB/BC, or the length of AD/DC, therefore, you can't tell.

14) DF < KH. There is a theorem that two triangles are congruent if two sides are congruent, and the angle between them (the included angles) are congruent in each of the triangles. A correlary (sp.?) to this theorem is that if the included angle of one triangle is greater than the included angle of the other triangle, so too is the opposite side. One way of thinking about this is to try exaggerating the diagram. Instead of having it be 100 degrees vs. 110 degrees, try making it 100 degrees vs. 170 degrees - the difference is seen easily, so you can pretty safely conclude that 110 vs 110 will not change much [if you keep the other two sides congruent].

15) AB = BC. Call the point on top of A point D, and the point below E. So, you have that line AD is parallel, and congruent to, line CE. Remember the theorem that alternate interior angles are congruent in parallel lines cut by a transversal? If not, trust me - it exists. If so, then that means that angle <ADB = angle <CEB. With this, you can prove that triangle ABD is congruent to triangle CEB, because you have two sides (AD = CE, DB = BE) are congruent, as are the angles between them (<ADB = < CEB). Since the two triangles are congruent, all the sides must be congruent to corresponding sides of the other triangle. Since AD = CE and DB = BE, the only remaining pair of angles is AB = BC. Another way to verify this is through the theorem that sides opposite congruent angles in congruent triangles are congruent, but this is a bit more confusing

16) Assumption: the opposite of what you are trying to prove; Assume: L || M.
Then: by alternate exterior angles are congruent in parallel lines cut by a transveral; Then: alt. exterior angles are congruent.
But: we are given that the alternate exterior angles 71 and 72 are NOT congruent.
So: there is a contradiction given our assumption, therefore our assumption must be false. L is NOT parallel to (can't draw symbol atm) M.

I hope I helped!! For further clarification, feel free to send me an MSN!

 

In what ways does Socrates suggest that sexual equality and the abolition of the family are necessary components of the &#8220;just city&#8221;? In what ways do these proposals illustrate the difficulties of achieving justice?

 

'
I'm gonna attempt this, however I'm sure the others will give a more clear answer, this is just off my head...

Socrates suggestion of sexual equality illustrates the difficulties of achieving justice because people have sexual bias all the time. For example, if potential employee A and B were of opposite sexes (A is male and B is female) and going for almost any job, we'll use trucking as an example, A is more likely to get employed if A and B were equal in all other ways because most sexually biased employers see the following;

A is a man, he will have more quick thinking, he won't have "girl problems" and will do man's will, get it done, now

B is a woman, she will not know what to do in a trucking emergency, as her thinking is not clear and concise, she must think about every option, which may take too long (even if it takes her less time to think of every possible option and pick one, than it takes the man to figure out the one. (I've seen examples of this).

This represents the supposed sexual inequality, and, unfortunately the opposite is true. If sexes were equal (they truly are), they should not make biased judgements based on this equality.

As for the family part of the argument, I have no secure opinion of it, so I will not give mine.

 

can you figure out the square root of pie? thats my homework

 

Lol. http://www.google.com/search?hl=en&q=sqrt+pi -- I sure hope you weren't serious, or had to like determine it in a semi-intelligent fashion...

Here's more info:

 

any idea on showing me the basic's on doing linear graphs?
im trying to help teach my little brother, sense i didnt have to learn this in high school idk how to do it.

 

Linear Graphs.. hmm.

y = mx + b

I believe that is the formula.

Let's say you have : y = (1/3)x + 6

m = slope
b = move up/down on Y-axis

I always suggest starting with B, because you'll have to move later.
-b = 6 in this case, so move up the Y-axis to 6, and make a point at (0,6)

In continuance, the slope it 1/3, which you use Rise over Run.
-Go up 1, and then over 3. In other words, move up 1 from 6, and then to the right 3.
-You should come up with another point - (3,7)

From here, draw a line through (0,6) and (3,7), and you have graphed a Linear Equation.

*NOTE*
If you need pictures for this, I would be more than happy to make some for you and post them.

 

What specifically about them?

y = mx + b is the emperical formula for ALL linear graphs. To find the equation of a linear line, all you need is two points...do you want to know how to do this? I mean I don't want to spend a few hours rambling on about linear graphing unless you have a relatively specific question. If you want the basics, check this out: http://en.wikipedia.org/wiki/Linear_equation

 

I have a paper due tomorrow on Antigone. Think you can help me write it or edit it?

Also, I have a D+ in chemistry right now, can you teach me stoichiometry?

 

Due tomorrow?!!? Could have posted this a bit earlier! That's sooo funny, since we just finished reading Antigone last week. What's the essay about??

 

Thats werid,Never mind though I finished it last night. It was about who was more of a tragic hero Creon or Antigone.

But I still need help with Chemistry. Can you teach me Stoichometry?

 

I'm sorry that we could not have responded sooner.
I also apologize that I know nothing about Stoichiometry, but you can find assistance here:
http://en.wikipedia.org/wiki/Stoichiometry

I had to drop my AP Chemistry class to retake Biology this year. Otherwise, I could help.

 

Need a little bit help please with my maths.

1)
cos(sin-1(x))=Ö(1-x2)
tan(sin-1(x))=x/Ö(1-x2)
What's the inversion?

2)
3·a+10·b180
5·a+10·b240


What's the coordinate?

3)

Is this right?

 

I got two questions

1.Every month, a girl gets allowance. Assume last year she had no money, and kept it up to now. Then she spends 1/2 of her money on clothes, then 1/3 of the remaining money on games, and then 1/4 of the remaining money on toys. After she bought all of that, she had $7777 left. Assuming she only gets money by allowance, how much money does she earn every month?

2.What is the sum of the coefficients of
( [3x - 3x^2 +1]^744 ) x ( [- 3x + 3x^2 +1]^745 )

 

Algebra baby
Let's say the amount of money she earns each month is called "x". At the start of this problem, since she kept it up to now, she will have 12x [She accumulates x dollars per month, 12 times].

Now, half of 12x = 6x. So, she spent "6x" on clothes. 1/3rd of 6x [12x - 6x, the remaining money after she spends money on clothes] is 2x, so she spent "2x' on clothes. Therefore, she has 12x - 6x - 2x = 4x after buying clothes and games. Then, she spends 1/4 of he remaining money (x, because 1/4 * 4x = x) on toys. So, she has spent 6x + 2x + x on "all that" and is left with (12x - 6x - 2x - x) or 3x.

So, 3x = 7777
x (the amount she earns per month) is 7777/3 = $2592.33

You can go back and check your work - it should work.

There is a rule of exponents;
(a^b) * (a^c) = a^(b+c).

So, since the stuff inside the parenthesis is the same (3x - 3x^2 + 1), this expression can be rewritten as:
(3x - 3x^2 + 1)^(744+745) = (3x - 3x^2 + 1)^1489.

The coefficients are 3 [from the 3x] and -3 [from the -3x^2], and 1 is a constant. I *think* the sum, therefore, is going to be 0. I could be way off, I'm going to have to re-check this. If it involves calculus/matrices (and/or Newton's Polynomial Expansion Theorem), I might be a bit stumped. If anyone wants to help, please pitch in because I'm almost sure I'm wrong.

Here are some links for you to check out, if you can understand it :O

http://en.wikipedia.org/wiki/Multinomial_theorem, http://www.trans4mind.com/personal_development/mathematics/series/multiNomialExpansion.htm