can someone please help me edit these essays? One explains my suspension and the other is a personal statement. They're for college. Thank you in advance.
During my sophomore year, I was in Mrs. Matsunaga’s French class. Mrs. Matsunaga and I have never got along very well. There are certain aspects of her teaching style that I dislike. For one, she speaks in a very quick manner. Furthermore when explaining concepts, she often opts for the difficult explanation, which confuses everybody. Yes she wants us to use proper terminology which is a good goal, but we won’t learn how it works if she doesn’t use a simple explanation first! There were also many things in French she mentioned that she didn’t want to explain because “We have to move on” or she doesn’t know about which is extremely frustrating, since personally, I like to have detailed explanations of topics we go over.
It was from this which originated my suspension. We were doing some type of writing activity, what exactly, I no longer recall but she had refused to explain something to me when the Spanish teacher walked in and began talking to her. I said “Fuuuuck, Mrs. Matsunaga.” A look of appall crossed her face. The next day Mrs. Matsunaga took me aside and said I had sworn at her. Despite my vehement denial, she would not believe me and I was sentenced to suspension for a day. Now it sounds like I did but I didn’t say “**** Mrs. Matsunaga.” I said “****, Mrs. Matsunaga” which was actually my method of expressing displeasure. Inappropriate and disrespectful? Yes. Swearing at her? No. To this day I do not understand why she will not believe me. Not only do I have no incentives to lie to her about it (I’ve already been suspended), I don’t have a past history of lying. Nonetheless, this has made me much more cautious in what I say, even in anger. Lashing out at people does not get a person very far, and it usually winds up hurting him or herself along with others. Better to try talking it out to try reach a solution instead of fuming and raging.
I'm still in French today. It wasn't very enjoyable, but I've weathered over these four years. To Mrs. Matsunaga, I speak with respect and tact which I've learned to use so that I don't get an instant defensive reaction from her. This is what I opt for nowadays too, instead of an argument, I try to have a discussion. I do not enjoy her class, but it has taught me something more valuable than all the French words I've ever learned: how to deal and work with people whom I don't get along with.
You might be better off tying it in to the topic language and pointing out that the whole convention of "swear words" is based on the linguistic snobbery of the French speaking Norman nobility against the Germanic words of Olde English. Hence "****", "****", etc., being curse words, whereas procreation, excrement, etc., are not.
Unless I'm horribly wrong somewhere (which is entirely possible since I don't remember integrals that well), if dt/dv = 1/(kv - 9.8), dv/dt = kv - 9.8. This derivative represents the rate of change of v wrt t. Integrating wrt t gets v(t) = t(kv - 9.8) + c, and since we know that v(0) = v_0, we get v(t) = t(kv - 9.8) + v_0. My further algebraic rearranging gets me v(t) = (-9.8t + v_0)/(1 - tk), though I don't know what you need.
I have come to seek help from the geometry gods.
Given:
AB=AC
Ad is perpendicular to BC, EF is perpendicular to AC
AF=6, ED=1
Find: The radius of circle E.
From what I can tell, AD is the angle bisector, median and altitude of A to triangle ABC. Additionally, triangle AFE is similar to both trinagles ABD and ADC through the AA similarity theorem. Feel free to make any necessary constructions, but avoid cos and sin. Also please explain your answer. Kthxbai.
Returning for more help: how do I prove that if 2 chords intersect inside a circle, the product of the lengths of the segments of 1 chord is = to the product of the lengths of the segments of the other chord? Just give a basic sumamry of the proof please.
Wow I poped in cause I usually help my son with homework and thought I could be helpful. Ha....Just looking at this page of posts I know I need to bow out of this forum instantly. Just reading those posts makes me feel dumb. Hope your homework comes out ok! Wow do I feel dumb! Ha!
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I think this is right, however I might be mistaken.
Real solutions:
First, ignore the absolute value. All that says is the graph will never be below the X axis.
Set the equation = to 0 ignoring the absolute value
(X^2)-4x-3 = 0
Use the quadratic Formula
(I hope you can do this because I don't feel like typing it)
You get -.645 and 4.645 as your x values
So assuming you were solving for X then these are your two answers.
The Graph will Look like a \/\/ Where the verticies only touch the X axis and do not pass it.
yeah - obviously I disagree with you, John. Spam Warning!.....Amazing - even in a forum where posts don't count, there's spam. That proves that getting rid of the post-count won't help getting rid of spam! Thans for this excellent example! - Craven
I think this is right, however I might be mistaken.
Real solutions:
First, ignore the absolute value.
I'm pretty sure you are mistaken (not a dig at you, an attempt to help the OP).
The absolute value is VERY important. |(x^2)-4x|=3 means that
1)(x^2)-4x = 3 (because |3| = 3)
OR
2) (x^2)-4x = -3 (because |-3| = 3)
Use the quadratic formula from here to derive the solutions for both subproblems/equations, and the global solution for your problem is the union of the solutions of both 1) and 2). (I'm expecting you can take it from here, if not drop me a PM and we'll go further on that problem.)
I haven't done the math, but it's likely from a quick back of the enveloppe look at it that you have four Real solutions here. Usually you'd have two for each of the sub equations, and I find it unlikely that they share a solution or that one of the equations only has one solution.
I'm pretty sure you are mistaken (not a dig at you, an attempt to help the OP).
The absolute value is VERY important. |(x^2)-4x|=3 means that
1)(x^2)-4x = 3 (because |3| = 3)
OR
2) (x^2)-4x = -3 (because |-3| = 3)
Use the quadratic formula from here to derive the solutions for both subproblems/equations, and the global solution for your problem is the union of the solutions of both 1) and 2). (I'm expecting you can take it from here, if not drop me a PM and we'll go further on that problem.)
I haven't done the math, but it's likely from a quick back of the enveloppe look at it that you have four Real solutions here. Usually you'd have two for each of the sub equations, and I find it unlikely that they share a solution or that one of the equations only has one solution.
Thank you very much. I got four answers 4.645, -.645, 3, 1. I plugged them back in and they all work leaving me with four real solutions. Thanks a bunch.
Find the value of a, such that polynomial, g(x)=(x^3)-(3ax^2)+12 is divisible by x+4
I came out with a = -19/12 but I don't know if that is the correct answer because it is a fraction.
Close, but no cigar. There are two ways to do this.
The easy way:
If g(x) is divisible by x+4 then g(4)=-48a-52=0 so a=-52/48=-13/12
The hard way:
Use long division
.......x^2-(3a+4)x+3
x+4) x^3-3ax^2+12
.....-(x^3+4x^2)
.....-(3a+4)x^2+12
.....-(-(3a+4)x^2-(3a+4)4x)
......(12a+16)x+12
.....-(3x+12)
......(12a+13)x=0
so a=-13/12
Can anyone help me with the problem 3sin120/6sin30-4cos45?
Here's my work, can anyone point out anything wrong?
3(1/2)/3-2√2 = 6/4 x 3 + 2√2 / 3 - 2√2 x 3 + 2√2 = 4.5 +3√2 / 1
= 4.5 + 3√2
Can anyone tell me if this is correct or not? I may have negative/positive signs mixed up, but if you come out with completely different numbers I have no idea what I'm doing wrong.
Let me first say that next time put in some parenthesis.
Instead of 3sin120/6sin30-4cos45, It would have looked better to say
3sin120/(6sin30-4cos45). It took me a while to realize that the 4cos45 was actually in the denominator.
Anyway, your work is very good I think I have found your mistake though. I believe that the value of sin120 is actually the (square root of 3) over (2).
Sqrt3/2, not 1/2
Which means in your first line: 3(1/2)/3-2√2, it should actually read...
3(Sqrt3/2)/3-2Sqrt2, if you rationalize the denominator from there, I found the solution to be 4.5 Sqrt3 + 3 Sqrt6. Which does check out with my calculator. Let me know if you need to see the work.
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Lets pull out the exponential then we get:
lim [e^x(1+x*e^-x)]^(1/x)=lim e(1+x*e^-x)^(1/x)=e*lim (1+x*e^-x)^(1/x)
In this case, the limit becomes 1^0 so the new limit is 1 and the solution is e.
Ok, let's take a break from math. I need some help. I need to find a monologue, from a courtroom TV show or movie, that is a closing argument, and that is between 1 and 2 minutes long. By tomorrow. I've been looking all week, and I haven't found squat. Anyone have any suggestions?
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Avvy: XenoNinja at HotP Studios
"Find the velocity of the object as a function of time if the initial velocity is V sub 0."
Thanks to the [Æther] shop for the sig!
You might be better off tying it in to the topic language and pointing out that the whole convention of "swear words" is based on the linguistic snobbery of the French speaking Norman nobility against the Germanic words of Olde English. Hence "****", "****", etc., being curse words, whereas procreation, excrement, etc., are not.
Find the velocity of the object as a function of time if the initial velocity is V sub 0, where dt/dv = 1/(kv - 9.8) and k is a constant.
Thanks to the [Æther] shop for the sig!
Given:
AB=AC
Ad is perpendicular to BC, EF is perpendicular to AC
AF=6, ED=1
Find: The radius of circle E.
From what I can tell, AD is the angle bisector, median and altitude of A to triangle ABC. Additionally, triangle AFE is similar to both trinagles ABD and ADC through the AA similarity theorem. Feel free to make any necessary constructions, but avoid cos and sin. Also please explain your answer. Kthxbai.
AE=BE=CE=R
ED^2+DC^2=1+DC^2=R^2 --> DC^2=R^2-1
AD^2+DC^2=AC^2=(AE+1)^2+DC^2=2R(R+1)
--> R=(-1+sqrt(1+2AC^2))/2
AF^2+EF^2=R^2 --> EF^2=R^2-36
EF^2+FC^2=R^2=(AC-6)^2+EF^2=(AC^2-12AC+36)+(R^2-36)
--> AC^2-12AC=0 --> AC=12
R=(-1+sqrt(1+2*12^2))/2=8
Thanks for the help, but how did you get 2R(R+1)?
Sorry, I do these kind things so quickly I often skip some steps.
(AE+1)^2+DC^2=(R+1)^2+DC^2=(R^2+2R+1)+(R^2-1)=2R^2+2R=2R(R+1)
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I thought about using the quadratic formula, but that whole absolute value thing is throwing of everything. How do I go about this?
Real solutions:
First, ignore the absolute value. All that says is the graph will never be below the X axis.
Set the equation = to 0 ignoring the absolute value
(X^2)-4x-3 = 0
Use the quadratic Formula
(I hope you can do this because I don't feel like typing it)
You get -.645 and 4.645 as your x values
So assuming you were solving for X then these are your two answers.
The Graph will Look like a \/\/ Where the verticies only touch the X axis and do not pass it.
So would the correcty answers be (-inf, -.646] U [4.65, inf) ?
Is zero itself allowed for abolute value or would I have to change the brackets to parenthesis?
Or x <= -.646 and x >= 4.65 ?
Same here is zero allowed for absolute or should I take out equal sign?
They're basically the same I think, but because the graph drops below axis I would have to use one of those correct?
I'm pretty sure you are mistaken (not a dig at you, an attempt to help the OP).
The absolute value is VERY important. |(x^2)-4x|=3 means that
1)(x^2)-4x = 3 (because |3| = 3)
OR
2) (x^2)-4x = -3 (because |-3| = 3)
Use the quadratic formula from here to derive the solutions for both subproblems/equations, and the global solution for your problem is the union of the solutions of both 1) and 2). (I'm expecting you can take it from here, if not drop me a PM and we'll go further on that problem.)
I haven't done the math, but it's likely from a quick back of the enveloppe look at it that you have four Real solutions here. Usually you'd have two for each of the sub equations, and I find it unlikely that they share a solution or that one of the equations only has one solution.
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Thank you very much. I got four answers 4.645, -.645, 3, 1. I plugged them back in and they all work leaving me with four real solutions. Thanks a bunch.
Find the value of a, such that polynomial, g(x)=(x^3)-(3ax^2)+12 is divisible by x+4
I came out with a = -19/12 but I don't know if that is the correct answer because it is a fraction.
Close, but no cigar. There are two ways to do this.
The easy way:
If g(x) is divisible by x+4 then g(4)=-48a-52=0 so a=-52/48=-13/12
The hard way:
Use long division
.......x^2-(3a+4)x+3
x+4) x^3-3ax^2+12
.....-(x^3+4x^2)
.....-(3a+4)x^2+12
.....-(-(3a+4)x^2-(3a+4)4x)
......(12a+16)x+12
.....-(3x+12)
......(12a+13)x=0
so a=-13/12
Here's my work, can anyone point out anything wrong?
3(1/2)/3-2√2 = 6/4 x 3 + 2√2 / 3 - 2√2 x 3 + 2√2 = 4.5 +3√2 / 1
= 4.5 + 3√2
Can anyone tell me if this is correct or not? I may have negative/positive signs mixed up, but if you come out with completely different numbers I have no idea what I'm doing wrong.
Instead of 3sin120/6sin30-4cos45, It would have looked better to say
3sin120/(6sin30-4cos45). It took me a while to realize that the 4cos45 was actually in the denominator.
Anyway, your work is very good I think I have found your mistake though. I believe that the value of sin120 is actually the (square root of 3) over (2).
Sqrt3/2, not 1/2
Which means in your first line: 3(1/2)/3-2√2, it should actually read...
3(Sqrt3/2)/3-2Sqrt2, if you rationalize the denominator from there, I found the solution to be 4.5 Sqrt3 + 3 Sqrt6. Which does check out with my calculator. Let me know if you need to see the work.
lim (e^x + x)^(1/x)
x-> inf
Well, it's an indeterminate form of type [inf]^0. L'Hopitale's won't go anywhere (this isn't an inference from the first sentence).
Just thinking about it nonrigorously, it seems like it should go to infinity.
Wait... couldn't you apply the result of the exponential function dwarfing every polynomial? With some kind of pinching approach?
Awesome avatar provided by Krashbot @ [Epic Graphics].
Lets pull out the exponential then we get:
lim [e^x(1+x*e^-x)]^(1/x)=lim e(1+x*e^-x)^(1/x)=e*lim (1+x*e^-x)^(1/x)
In this case, the limit becomes 1^0 so the new limit is 1 and the solution is e.
Many thanks to:
Sig: CharlieD at Limited Edition Signatures
Avvy: XenoNinja at HotP Studios
I cant work with a calculator so showing me the calculator signs doesnt help me at all.
Basically im just simplifying the radicals.
Attached are the problems and the answers that follow them (the answers are behind the book im using).
If some one could show me (it helps me more) and not tell how to do them it would be very helpful.
thanks in advance.
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