I wasn't joking at all. Seriously, the pythagorean theorem is more work than you need to do. All you have to do is see if the line segments are perpendicular.
MDay
Diffentiate such simple function might not be the quickest way to solve this problem as you can convert the equations to y=mx+b format STRAIGHT away knowing M is the slope and B is the intercept already.
For harder functions, differentiation is the way to go. But not such simple, easily rearranged functions.
Diffentiate such simple function might not be the quickest way to solve this problem as you can convert the equations to y=mx+b format STRAIGHT away knowing M is the slope and B is the intercept already.
For harder functions, differentiation is the way to go. But not such simple, easily rearranged functions.
let's say the slope of one line segment f(x) is a and the slope of another line segment g(x)is b
then:
f(x) and g(x) are perpendicular if and only if, a = -1/b.
for 2 perpendicular lines, slope of one is 1/2 (one half) the other must be -2 (negative 2)
then:
f(x) and g(x) are perpendicular if and only if, a = -1/b.
for 2 perpendicular lines, slope of one is 1/2 (one half) the other must be -2 (negative 2)
xtreme2k,
i was kidding with the differentiation, duh.
the kid, urbantechie, does not even know calculus ;-)
besides, with IMPLICIT differentiation, it's easy ;-)
and why must you misspell my nick?
i was kidding with the differentiation, duh.
the kid, urbantechie, does not even know calculus ;-)
besides, with IMPLICIT differentiation, it's easy ;-)
and why must you misspell my nick?
<< welcome to the official AT HOMEWORK Help Thread. What is your problem >>
okay, i'll bite
this is my math problem. it's from algebra 1 ;-)
Prove that if K is any field which containes D then K contains a subfield isomorphic to F. (In this sense F is the smallest field containing D.)
assume that D is an integral domain.
mAdD INDIAN
I think with your question, there is no need to do it like that.
For a triangle to be inscribed by a SEMI circle, with BC being the DIAMETER of the circle, the OPPOSITE angle is always 90' (I think this is circle geometry - hope I am correct)
IF that opposite angle must be 90' then the whole things becomes easy.
Now you want the max HEIGHT of the trangle. Since the area of a triangle is alwas base*height. Base you cannot change, therefore you want maximum height.
But the max HEIGHT of the circle must be 5 cm (radius of the circle)
Therefore, 5cm = 5sin@
@ must be 45'
I think with your question, there is no need to do it like that.
For a triangle to be inscribed by a SEMI circle, with BC being the DIAMETER of the circle, the OPPOSITE angle is always 90' (I think this is circle geometry - hope I am correct)
IF that opposite angle must be 90' then the whole things becomes easy.
Now you want the max HEIGHT of the trangle. Since the area of a triangle is alwas base*height. Base you cannot change, therefore you want maximum height.
But the max HEIGHT of the circle must be 5 cm (radius of the circle)
Therefore, 5cm = 5sin@
@ must be 45'
man i am stuck in this one 
The amplitude of vibration, A is given by
A = F / (c^2 * m^2 + (k-mw^2)^2)^(1/2)
1. calculate @A/@c and @A/@k <- i can do this part (@ means partial diff)
2. if the maximum error in the experimentally determined values of c and k are 5% each, show that the maximum % error in the predicted amplitude of vibration is less than 5 ((c^2 * m^2 + k(k-mw^2))/(c^2 m^2 + (k - mw^2)^2))
The amplitude of vibration, A is given by
A = F / (c^2 * m^2 + (k-mw^2)^2)^(1/2)
1. calculate @A/@c and @A/@k <- i can do this part (@ means partial diff)
2. if the maximum error in the experimentally determined values of c and k are 5% each, show that the maximum % error in the predicted amplitude of vibration is less than 5 ((c^2 * m^2 + k(k-mw^2))/(c^2 m^2 + (k - mw^2)^2))
Thank you Xtreme2K.
I kinda figured that Ø = 45 but I didn't know how to prove it so I thought of doing it using differentiation.
Thanks again..I'll try going over it again.
I have another problem for you guys (if you have time and don't mind answering).
f(x)= x * sin(x) + cos(x) , -pi < x < pi ---> find local max/min points.
I know the derivative of f(x) is ::
f'(x) = x * cos(x)
now how do I find the critical number of that?
I kinda figured that Ø = 45 but I didn't know how to prove it so I thought of doing it using differentiation.
Thanks again..I'll try going over it again.
I have another problem for you guys (if you have time and don't mind answering).
f(x)= x * sin(x) + cos(x) , -pi < x < pi ---> find local max/min points.
I know the derivative of f(x) is ::
f'(x) = x * cos(x)
now how do I find the critical number of that?
mAdD INDIAN
f(x) = x * sin x + cos x
f'(x) = x * cos x + 1 sin x - sin x
f'(x) = x * cos x
you want a STATIONARY Point f'(x) = 0
x * cos x = 0
either
x = 0 OR cos x = 0
therefore
x = 0 OR pi/4, 3pi/4, -pi/4, -3pi/4
now we need the 2nd devirative to see which point is max/min
... :< i am starting to get a bit rusty
f(x) = x * sin x + cos x
f'(x) = x * cos x + 1 sin x - sin x
f'(x) = x * cos x
you want a STATIONARY Point f'(x) = 0
x * cos x = 0
either
x = 0 OR cos x = 0
therefore
x = 0 OR pi/4, 3pi/4, -pi/4, -3pi/4
now we need the 2nd devirative to see which point is max/min
... :< i am starting to get a bit rusty
someone help me i'm gonna fail my linear algebra final..... ack. and those geometry problems were pathetic, you shouldn't need help in high school, until calc at the worst, its all way too easy.
Honestly I dont know anyone that still needs tutoring at uni
I mean, if you need tutoring at uni, you are in the wrong course basically
I mean, if you need tutoring at uni, you are in the wrong course basically
This brings back memories...
Aren't two lines perpendicular if their slopes are the inverse of each other?
Like:
y = x
y = -x
Are perpendicular? The slope of the first being 1, the slope of the second being -1.. Seems like the quick and easy way to tell.
Aren't two lines perpendicular if their slopes are the inverse of each other?
Like:
y = x
y = -x
Are perpendicular? The slope of the first being 1, the slope of the second being -1.. Seems like the quick and easy way to tell.
perry
for 2 straight lines to be perpendicular, their gradient must be -1 when you multiply them.
In another words if a is the gradient of line A and b is the gradient of line B, these lines are perpendicular ONLY if
a * b = -1
for 2 straight lines to be perpendicular, their gradient must be -1 when you multiply them.
In another words if a is the gradient of line A and b is the gradient of line B, these lines are perpendicular ONLY if
a * b = -1
I have a C++ project to do and it's all about arrays. Something about turtle graphics. If someone here, whose VERY knowleadgeable in C++ is willing to help then I'd appreicate it This program would probably take you 5 minutes to write and me an eternity because I am behind in the class. There will be a little reward in it for ya if you're willing to "help" if ya know what I mean
I now declare this thread "the help me cheat" thread. E-mail me at aaron_van@msn.com
This program would probably take you 5 minutes to write and me an eternity because I am behind in the class. There will be a little reward in it for ya if you're willing to "help" if ya know what I mean I now declare this thread "the help me cheat" thread. E-mail me at aaron_van@msn.com
The proof is like this i think
m1 and m2 are gradient of line 1 and 2
@ is the angle of intersection of 2 lines
tan @ = (m1 - m2)/1 + m1*m2
since @ must be 90' (perpendicular) there fore tan @ is infinity
for tan @ to be infinity
1 + m1*m2 must be 0
1 + m1 * m2 = 0
m1 * m2 = -1
m1 and m2 are gradient of line 1 and 2
@ is the angle of intersection of 2 lines
tan @ = (m1 - m2)/1 + m1*m2
since @ must be 90' (perpendicular) there fore tan @ is infinity
for tan @ to be infinity
1 + m1*m2 must be 0
1 + m1 * m2 = 0
m1 * m2 = -1
OK, I posted this in the software forum but haven't got a reply so here goes.
This is REALLY basic stuff for you Java gurus, but I'm just learning. I have to do this problem for my programming class and here is the problem as it appears in my book:
<< You can change the size of an applet by using the setSize(int h, int v) method. Write an applet that contains two Buttons, labeled "Big" and "Small." Whenever the user clicks on small, set the applet's dimensions to 200 x 100, and whenever the user clicks on big, set the dimensions to 300 x 200. >>
And this is the code that I have so far:
<<
import java.applet.*;
import java.awt.*;
import java.awt.event.*;
public class SimpleApplet extends Applet implements ActionListener
{
private Button changeBig;
private Button changeSmall;
public void init()
{
changeBig = new Button("Big"
changeBig.addActionListener(this);
changeSmall = new Button("Small"
changeSmall.addActionListener(this);
add(changeBig);
add(changeSmall);
setSize(300, 200);
}
public void actionPerformed(ActionEvent e)
{
if (e.getSource() == changeBig)
setSize(300,200);
else if (e.getSource() == changeSmall)
setSize(200, 100);
}
} >>
I'm using JBuilder 3 University edition to compile. When I run this in Internet Explorer 5.5, it gives me a grey box with the two buttons towards the top, but clicking on them does nothing. I don't think the setSize() method is doing anything though, because the initial setSize() under the init() doesn't even work. No matter what I change that value to, the applet is always the default size when it starts up, then pressing the buttons does nothing. What am I doing wrong here?
This is REALLY basic stuff for you Java gurus, but I'm just learning. I have to do this problem for my programming class and here is the problem as it appears in my book:
<< You can change the size of an applet by using the setSize(int h, int v) method. Write an applet that contains two Buttons, labeled "Big" and "Small." Whenever the user clicks on small, set the applet's dimensions to 200 x 100, and whenever the user clicks on big, set the dimensions to 300 x 200. >>
And this is the code that I have so far:
<<
import java.applet.*;
import java.awt.*;
import java.awt.event.*;
public class SimpleApplet extends Applet implements ActionListener
{
private Button changeBig;
private Button changeSmall;
public void init()
{
changeBig = new Button("Big"
;changeBig.addActionListener(this);
changeSmall = new Button("Small"
;changeSmall.addActionListener(this);
add(changeBig);
add(changeSmall);
setSize(300, 200);
}
public void actionPerformed(ActionEvent e)
{
if (e.getSource() == changeBig)
setSize(300,200);
else if (e.getSource() == changeSmall)
setSize(200, 100);
}
} >>
I'm using JBuilder 3 University edition to compile. When I run this in Internet Explorer 5.5, it gives me a grey box with the two buttons towards the top, but clicking on them does nothing. I don't think the setSize() method is doing anything though, because the initial setSize() under the init() doesn't even work. No matter what I change that value to, the applet is always the default size when it starts up, then pressing the buttons does nothing. What am I doing wrong here?
I hope you guys caught the error that zippy made:
Y+6=1/2(x-4)
Y+6=1/2x-2
~-6~~~~ -6
Y=1/2x-8
m=1/2, b=8
Since the formula is Y=mx+b, b is -8. He got the second one right though and it's the same thing. I'm glad I'm done with all the math I could ever use... =)
Y+6=1/2(x-4)
Y+6=1/2x-2
~-6~~~~ -6
Y=1/2x-8
m=1/2, b=8
Since the formula is Y=mx+b, b is -8. He got the second one right though and it's the same thing. I'm glad I'm done with all the math I could ever use... =)
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