Can i get C Exercises on net?As i have completed till Nested if-else and loops and there is no sufficient problems available in the book, i m not able to practice well .Now i m searching for site where i can get some good C Exercise to built my basic learning ?
Can some tell me sites whihc provide Basic C Exercises and also tell some books which is exclusivly used for problem solving for basic C?
Thanks
C Exercises?
I feel you shd try some forum and yahoo groups.There you will get some good stuff .... and help too
Reply:Or else you may contact a C expert live at website like http://oktutorial.com/ .
Reply:The best way to get practise is to do the problems that people post right here.
Like this simple one I answered with recursion but you may want to try to rewrite it without recursion.
http://answers.yahoo.com/question/index;...
Monday, May 24, 2010
C-section moms?
I am due to have a c-section at 39 weeks. I am 35 weeks and 4 days right now. I went in two days ago with contractions every 3 min but I was only dialted to 1cm and stayed that way for a couple hours so they stoped the contractions with a shot. Ever sice that night I still have contractions and I have droped so much that everyone noticed. My question is do I go to the doctors again or do I wait until I am in complete active labor? I don't want the medication again and I don't want to wait so long that I have to have an emercency c-section. The nurse said at this point they will not try to stop active labor and if I was dialted more they would just do the c-section. My next appointment is 1 week away. I know normal births need to wait for contractions to be close and strong but for c-sections and early labor it is not the same.
C-section moms?
i would go to the doctor or to the hospital if i were you. i was1cm dilated then 5hrs later i was 7! i wasn't meant to have a c section but my baby turned transverse breech. if you are having a cesarean they would prefer your waters not to break. if its becoming active labour you must go to the hospital. it's better off to get checked out. i was 1 cm dilated for a week then it went full scale. they put you in if your in pain that you can't handle and if you can't sleep . i had a sleeping pill cause i did't sleep for a week. so it's better off they can keep an eye on you.
Reply:My friend was scheduled to have a c-section and actually her water broke and had to have the baby early! like almost 2 weeks! she was fine but was in NICU for a day.
Reply:If the contractions come and go (maybe for an hour each day), then I'd wait it out. However, if you've had contractions since then, I'd go in to get checked. Which is worse - getting a shot or having a problem with your child and creating an emergency situation since you didn't want to go in?
Besides, at 36 weeks, they don't stop labor and chances are your child will require minimal NICU time - lungs are totally fine by 36-37 weeks. So, if you're more dialated, it may just be your body's way of letting you know this baby is ready to be born.
C-section moms?
i would go to the doctor or to the hospital if i were you. i was1cm dilated then 5hrs later i was 7! i wasn't meant to have a c section but my baby turned transverse breech. if you are having a cesarean they would prefer your waters not to break. if its becoming active labour you must go to the hospital. it's better off to get checked out. i was 1 cm dilated for a week then it went full scale. they put you in if your in pain that you can't handle and if you can't sleep . i had a sleeping pill cause i did't sleep for a week. so it's better off they can keep an eye on you.
Reply:My friend was scheduled to have a c-section and actually her water broke and had to have the baby early! like almost 2 weeks! she was fine but was in NICU for a day.
Reply:If the contractions come and go (maybe for an hour each day), then I'd wait it out. However, if you've had contractions since then, I'd go in to get checked. Which is worse - getting a shot or having a problem with your child and creating an emergency situation since you didn't want to go in?
Besides, at 36 weeks, they don't stop labor and chances are your child will require minimal NICU time - lungs are totally fine by 36-37 weeks. So, if you're more dialated, it may just be your body's way of letting you know this baby is ready to be born.
(b-a)(c-a)(c-b)=what matrix?
(b-a)(c-a)(c-b)=1 1 1
a b c
a^2 b^2 c^2
prove this matrix
(b-a)(c-a)(c-b)=what matrix?
Ideas: Subtract the values of the first column from the second column and the third column, respectively.
So, the determinant
= (b-a)(c^2 - a^2) - (c-a)(b^2 - a^2)
= (b-a)(c-a)(c+a - b-a)
= (b-a)(c-a)(c-b)
Reply:I do not understand the question.
b-a=1
b=a
c-a=1
c=a
c-b=1
c=b
a=b=c=1
a b c = 1 1 1
a^2 b^2 c^2 = 1 1 1
So the Matrix is
1 1 1
1 1 1
1 1 1
Is this what you wanted to prove? The matrix should have 3 rows and 3 columns.
rose
a b c
a^2 b^2 c^2
prove this matrix
(b-a)(c-a)(c-b)=what matrix?
Ideas: Subtract the values of the first column from the second column and the third column, respectively.
So, the determinant
= (b-a)(c^2 - a^2) - (c-a)(b^2 - a^2)
= (b-a)(c-a)(c+a - b-a)
= (b-a)(c-a)(c-b)
Reply:I do not understand the question.
b-a=1
b=a
c-a=1
c=a
c-b=1
c=b
a=b=c=1
a b c = 1 1 1
a^2 b^2 c^2 = 1 1 1
So the Matrix is
1 1 1
1 1 1
1 1 1
Is this what you wanted to prove? The matrix should have 3 rows and 3 columns.
rose
Compatability between C/C++ compilers?
I'm wondering whether C source code can be compiled in a C++ compiler, or do you have to use a C compiler.
Also, does anyone have any recommendations on a good C compiler?
Thanks!!
Compatability between C/C++ compilers?
They are usually compatible if you stick to ANSI C/C++ coding. A good compile is the free gnu C/C++ compiler. This is a well documented stable compiler that runs on windows, linux etc..
Reply:I wrote numerous programs using Watcom C/C++, now free:
http://www.openwatcom.org/index.php/Main...
Most compile with .C and .CPP names.
Also, does anyone have any recommendations on a good C compiler?
Thanks!!
Compatability between C/C++ compilers?
They are usually compatible if you stick to ANSI C/C++ coding. A good compile is the free gnu C/C++ compiler. This is a well documented stable compiler that runs on windows, linux etc..
Reply:I wrote numerous programs using Watcom C/C++, now free:
http://www.openwatcom.org/index.php/Main...
Most compile with .C and .CPP names.
Compatability between C/C++ compilers?
I'm wondering whether C source code can be compiled in a C++ compiler, or do you have to use a C compiler.
Also, does anyone have any recommendations on a good C compiler?
Thanks!!
Compatability between C/C++ compilers?
They are usually compatible if you stick to ANSI C/C++ coding. A good compile is the free gnu C/C++ compiler. This is a well documented stable compiler that runs on windows, linux etc..
Reply:I wrote numerous programs using Watcom C/C++, now free:
http://www.openwatcom.org/index.php/Main...
Most compile with .C and .CPP names.
Also, does anyone have any recommendations on a good C compiler?
Thanks!!
Compatability between C/C++ compilers?
They are usually compatible if you stick to ANSI C/C++ coding. A good compile is the free gnu C/C++ compiler. This is a well documented stable compiler that runs on windows, linux etc..
Reply:I wrote numerous programs using Watcom C/C++, now free:
http://www.openwatcom.org/index.php/Main...
Most compile with .C and .CPP names.
The Circle C has the equation x^2 + y^2 - 12x + 8y +16 = 0?
a) Find the coordinates of the centre of C
b) Find the radius of C
c) Sketch C
Given that C crosses the x-axis at the points A and B
d) Find the length AB, giving your answer in the form K√5
Its the last question i'm stuck on for my homework, any help would be great :)
The Circle C has the equation x^2 + y^2 - 12x + 8y +16 = 0?
a) General equation of a circle can be expressed in
x^2 +y^2 +2gx+2fy +c = 0 with (-g,-f) as centre and
SQRT(g^2 + f^2 -c)
=%26gt; Hence the above equation can be transformed to
x^2 + y^2 + 2(-6)x + 2(4)y + 16 = 0 ,
i.e g = -6 , f = 4 , hence the centre of the above circle is (6,-4) .
b) Radius = SQRT(6^2 + (-4)^2 - 16)) = SQRT(36) = 6 Units
c) You can draw the Circle with (6,-4) as centre and 6 as radius on a coordinate axes.
d) For solving this , put y = 0 in the equation of the circle and solve for point A %26amp; B.
=%26gt; x^2 + 0^2 - 12x + 8x0 + 16 = 0
=%26gt; x^2 -12x + 16 = 0
Solving for x , you will get the coordinates for A and B as follows:A(6+2√5,0) and B(6-2√5,0). Hence the length of AB
= 4√5 units
Reply:We need to put this in the form (x-a)^2 + (b-y)^2 = r^2, and we get (x-6)^2 -36 + (y+4)^2 -16 +16 = 0. This simplifies to (x-6)^2 + (y+4)^2 = 36. This takes care of parts a and b. Substitute y = 0 in this, solve for the two values of x, take their difference, and you're done.
Reply:Perhaps this may help:
The center of a circle is defined as (h,k)
h is the x value, k is the y value
in order to find (h,k) you have to get the equation into the form
(x-h)^2 + (y-k)^2 = r^2 r here is the radius
to put your equation in to the right form you must group the x values and the y values.
so you get: x^2 -12x + y^2 + 8y= - 16
(note that the 16 was brought over to the other side)
now lets work with x and y separately
right now we have x^2 - 12X
we must complete the square. take the -12 and cut it in half.
we now have 6. square this number to get 36
[Make sure you add the 36 to the -16 on the other side of the equal sign!]
so now you have x^2-12x+36
Factor this to get (x-6)^2
now do the same for the y: cut the 8y in half, square it, add that value to the -16 on the other side, and factor.
this should lead you to (y+4)^2
now all in all you should have (x-6)^2 + (y+4)^2 = 36
(the 36 comes from [ -16+36+16]
so looking at this equation we can tell that the center of the circle is (6, -4)
to find the radius we remember that (x-h)^2+(y-k)^2=r^2
in this case r^2 = 36 so r equals the square root of 36 which equals six.
I hope this helps you with the first questions of your homework and should you need further help you might find it at http://www.analyzemath.com/CircleEq/Tuto...
Reply:(x-6)^2 + (y+4)^2 = 36+16-16 = 36
[(x-6)/6]^2 + [(y+4)/6]^2 = 1
origin (6,-4)
radious 6
Reply:center is(-a/2,-b/2)=(12/2,-8/2)=(6,-4)
Reply:As many have said,
(x - 6)^2 + (y + 4)^2 = 36.
The x-axis is y=0, so the two x-values are the solutions of
(x - 6)^2 + (0 + 4)^2 = 36;
i.e.
(x-6)^2 = 20; x = 6 ± √20 = 6 ± 2√5.
The distance between the two is
(6 + 2√5) - (6 - 2√5) = 4√5
Reply:First of all, complete the square to get the equation in the form
(x - x0)^2 + (y - y0)^2 = r^2
(x^2 - 12x) + (y^2 + 8y) = -16
(x^2 - 12x + 36) + (y^2 + 8y + 16) = -16 + 36 + 16
(x - 6)^2 + (y + 4)^2 = 6^2
a) From the equation above, x0 = 6 and y0 = -4, so the coordinates of the center are (6, -4).
b) Again, from the equation above, r = 6.
c) Simply draw a circle centered at the point (6, -4) with a radius of 6 units.
d) I'm not sure what you mean by AB. More information is needed.
Reply:x^2 - 12x + y^2 + 8y = -16.
(x - 6)^2 + (y + 4)^2 = -16 + 36 + 16
(x - 6)^2 + (y + 4)^2 = 36.
A) Center is located at (6, -4).
B) The radius is 36^.5 = 6 units.
D) I'd help you with the length of AB, but you didn't even mention A or B in your answer.
b) Find the radius of C
c) Sketch C
Given that C crosses the x-axis at the points A and B
d) Find the length AB, giving your answer in the form K√5
Its the last question i'm stuck on for my homework, any help would be great :)
The Circle C has the equation x^2 + y^2 - 12x + 8y +16 = 0?
a) General equation of a circle can be expressed in
x^2 +y^2 +2gx+2fy +c = 0 with (-g,-f) as centre and
SQRT(g^2 + f^2 -c)
=%26gt; Hence the above equation can be transformed to
x^2 + y^2 + 2(-6)x + 2(4)y + 16 = 0 ,
i.e g = -6 , f = 4 , hence the centre of the above circle is (6,-4) .
b) Radius = SQRT(6^2 + (-4)^2 - 16)) = SQRT(36) = 6 Units
c) You can draw the Circle with (6,-4) as centre and 6 as radius on a coordinate axes.
d) For solving this , put y = 0 in the equation of the circle and solve for point A %26amp; B.
=%26gt; x^2 + 0^2 - 12x + 8x0 + 16 = 0
=%26gt; x^2 -12x + 16 = 0
Solving for x , you will get the coordinates for A and B as follows:A(6+2√5,0) and B(6-2√5,0). Hence the length of AB
= 4√5 units
Reply:We need to put this in the form (x-a)^2 + (b-y)^2 = r^2, and we get (x-6)^2 -36 + (y+4)^2 -16 +16 = 0. This simplifies to (x-6)^2 + (y+4)^2 = 36. This takes care of parts a and b. Substitute y = 0 in this, solve for the two values of x, take their difference, and you're done.
Reply:Perhaps this may help:
The center of a circle is defined as (h,k)
h is the x value, k is the y value
in order to find (h,k) you have to get the equation into the form
(x-h)^2 + (y-k)^2 = r^2 r here is the radius
to put your equation in to the right form you must group the x values and the y values.
so you get: x^2 -12x + y^2 + 8y= - 16
(note that the 16 was brought over to the other side)
now lets work with x and y separately
right now we have x^2 - 12X
we must complete the square. take the -12 and cut it in half.
we now have 6. square this number to get 36
[Make sure you add the 36 to the -16 on the other side of the equal sign!]
so now you have x^2-12x+36
Factor this to get (x-6)^2
now do the same for the y: cut the 8y in half, square it, add that value to the -16 on the other side, and factor.
this should lead you to (y+4)^2
now all in all you should have (x-6)^2 + (y+4)^2 = 36
(the 36 comes from [ -16+36+16]
so looking at this equation we can tell that the center of the circle is (6, -4)
to find the radius we remember that (x-h)^2+(y-k)^2=r^2
in this case r^2 = 36 so r equals the square root of 36 which equals six.
I hope this helps you with the first questions of your homework and should you need further help you might find it at http://www.analyzemath.com/CircleEq/Tuto...
Reply:(x-6)^2 + (y+4)^2 = 36+16-16 = 36
[(x-6)/6]^2 + [(y+4)/6]^2 = 1
origin (6,-4)
radious 6
Reply:center is(-a/2,-b/2)=(12/2,-8/2)=(6,-4)
Reply:As many have said,
(x - 6)^2 + (y + 4)^2 = 36.
The x-axis is y=0, so the two x-values are the solutions of
(x - 6)^2 + (0 + 4)^2 = 36;
i.e.
(x-6)^2 = 20; x = 6 ± √20 = 6 ± 2√5.
The distance between the two is
(6 + 2√5) - (6 - 2√5) = 4√5
Reply:First of all, complete the square to get the equation in the form
(x - x0)^2 + (y - y0)^2 = r^2
(x^2 - 12x) + (y^2 + 8y) = -16
(x^2 - 12x + 36) + (y^2 + 8y + 16) = -16 + 36 + 16
(x - 6)^2 + (y + 4)^2 = 6^2
a) From the equation above, x0 = 6 and y0 = -4, so the coordinates of the center are (6, -4).
b) Again, from the equation above, r = 6.
c) Simply draw a circle centered at the point (6, -4) with a radius of 6 units.
d) I'm not sure what you mean by AB. More information is needed.
Reply:x^2 - 12x + y^2 + 8y = -16.
(x - 6)^2 + (y + 4)^2 = -16 + 36 + 16
(x - 6)^2 + (y + 4)^2 = 36.
A) Center is located at (6, -4).
B) The radius is 36^.5 = 6 units.
D) I'd help you with the length of AB, but you didn't even mention A or B in your answer.
C-section the 2nd time around?
I had my baby 6 months ago and am just wondering if when I have a baby again(eventually) will I have to have another c-section or can i have a natural birth? I didnt mind having a c-section last time(it wasnt by choice) but in the Australian public system can you choose to have a c-section or is it only for emergencies? I have heard if you do it naturally after having a c-section 1st time around you can split your scar open?????
C-section the 2nd time around?
Yes, you can have your next baby via VBAC (vaginal birth after cesarean) as long as everything is medically ok. I had an emergency c-section with my first child and wanted to try a VBAC with my second, but my doctor warned me that it would not be safe considering how they made the incision with my first. Yes, there is a chance of splitting at your previous incision. I'm not sure how the Australian system works, but in the U.S., a woman can choose to have a scheduled c-section (not that it's always the best choice when done for the wrong reasons).
flowering plum
C-section the 2nd time around?
Yes, you can have your next baby via VBAC (vaginal birth after cesarean) as long as everything is medically ok. I had an emergency c-section with my first child and wanted to try a VBAC with my second, but my doctor warned me that it would not be safe considering how they made the incision with my first. Yes, there is a chance of splitting at your previous incision. I'm not sure how the Australian system works, but in the U.S., a woman can choose to have a scheduled c-section (not that it's always the best choice when done for the wrong reasons).
flowering plum
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