a.wireframes
b.autocads
c.pixelplots
d.bitmaps
The two basic groups of graphic file formats are vectors and?
bitmaps
Reply:I onl know the answer in Dutch, but kind of translating it to English, I would say c: Pixelplots (If I translate it exactly from Dutch to English, I would say grid-format)
Reply:bitmaps, which refers to the pattern of labelling the bits in a raster scan from top to bottom
Reply:Take the damn test yourself! This is not testing your knowledge
Reply:I would have said:
a: wireframes - to tie you with for being so silly
b: autocads - but you ride cars, not "cads"
c: pixelplots - but then the size of your brain isnt even up to a pixel
d. bitmaps - But I'm sure you dont know one-bit aboout maps
But what the heck! I'm sure you just read it up somewhere last night, and you want to make a public show of your ignorance to the world!
but sorry, nobody is goning to answer your silly question.
Friday, May 21, 2010
Vector Componants Please help!?
Which of following sets of conditions (A - F), if true, would show that the expressions 1 and 2 above define the same vector as expressions 3 and 4?
1) C= Cosine Law Equation
2) o/ = sin^-1 (Bsin(c) / C )
3) Cx = A + B Cos(theta)
4) Cy = Bsin(theta)
a)The two pairs of expressions give the same length and direction for Vector C.
b)The two pairs of expressions give the same length and x component for Vector C.
c)The two pairs of expressions give the same direction and x component for Vector C.
d)The two pairs of expressions give the same length and y component for Vector C.
e)The two pairs of expressions give the same direction and y component for Vector C.
f)The two pairs of expressions give the same x and y components for Vector C.
Vector Componants Please help!?
Answer: E
The two pairs of expressions give the same direction and y component for Vector C.
sending flowers
1) C= Cosine Law Equation
2) o/ = sin^-1 (Bsin(c) / C )
3) Cx = A + B Cos(theta)
4) Cy = Bsin(theta)
a)The two pairs of expressions give the same length and direction for Vector C.
b)The two pairs of expressions give the same length and x component for Vector C.
c)The two pairs of expressions give the same direction and x component for Vector C.
d)The two pairs of expressions give the same length and y component for Vector C.
e)The two pairs of expressions give the same direction and y component for Vector C.
f)The two pairs of expressions give the same x and y components for Vector C.
Vector Componants Please help!?
Answer: E
The two pairs of expressions give the same direction and y component for Vector C.
sending flowers
Someone who inconspicuously harbors a pathogen and spreads it to others is a?
a. fomite
b. carrier
c. vector
d. reservoir
e. source
Someone who inconspicuously harbors a pathogen and spreads it to others is a?
c
Reply:Vector
Also, Typhoid Mary.
Reply:THEY ARE KNOWN AS CARRIERS. ONE OF THE MOST FAMOUS WAS TYPHOID MARY.
Reply:a carrier i believe because they may not have the disase themselves but they would be able to transfer it
b. carrier
c. vector
d. reservoir
e. source
Someone who inconspicuously harbors a pathogen and spreads it to others is a?
c
Reply:Vector
Also, Typhoid Mary.
Reply:THEY ARE KNOWN AS CARRIERS. ONE OF THE MOST FAMOUS WAS TYPHOID MARY.
Reply:a carrier i believe because they may not have the disase themselves but they would be able to transfer it
C++ help, can you help??
I need to write a function that adds the elements in two vectors, the vectors are arrays this my function defintion thus far:
//code
void vectorSum(double vecSum[], int%26amp; size){
double vec1;
double vec2;
for(int i= 0; i %26lt; size; i++){
vecSum[i]= (vec1[i] + vec2[i]);
}
C++ help, can you help??
What you miss is that both the input vectors and the output vectors must be parameters to the function. By making them parameters to the function, you tell the function what two vectors to add and where to put the sum. Also, you don't need to make size a reference.
You want something like this:
void vectorSum(double vecSum[], double const vec1[], double const vec2[], int size){
for(int i= 0; i %26lt; size; i++){
vecSum[i]= (vec1[i] + vec2[i]);
}
Now you can use your function like this:
int const v_size = 20;
int vec1[v_size];
int vec2[v_size];
int vec_sum[v_size];
// fill vec1 and vec2 with values here
vectorSum(vec_sum, vec1, vec2, v_size);
// Now vec_sum is the sum of vec2 and vec2
Reply:Some quick thoughts:
1. vec1 and vec2 haven't been defined properly.
2. vec1 and vec2 haven't been given any values.
3. Why are vec1 and vec2 inside the function? Shouldn't they be outside it?
Hope that helps.
Reply:Hello. You don't declare vec1 and vec2 as arrays. I'm just a tired old C programmer up past his bed time but it looks as if you declare them as doubles, that is as single variables, not arrays of variables, and certainly do not initialize them. If they are supposed to be global variables, why do you declare them in the block? And why aren't they arrays if you're going to treat them as arrays? If I were a good compiler, I'd assign a space for a double called vec1 and next to it a space for a double called vec2. Then if I were a good but nasty compiler, I would on the vecSum[i] line add vec1 and vec2 for 0, vec2 and whatever garbage is next to it for 1, the garbage I used for vec2[1] for the new vec1[2] and the garbage next to that...
More likely your compiler isn't nasty so it's just declaring an error and refusing to compile.
You could possibly declare them pointers and in C terms malloc a memory length of size*sizeof(double) for each. Actually you just shouldn't declare them in this function. You should pass them as parameters.
//code
void vectorSum(double vecSum[], int%26amp; size){
double vec1;
double vec2;
for(int i= 0; i %26lt; size; i++){
vecSum[i]= (vec1[i] + vec2[i]);
}
C++ help, can you help??
What you miss is that both the input vectors and the output vectors must be parameters to the function. By making them parameters to the function, you tell the function what two vectors to add and where to put the sum. Also, you don't need to make size a reference.
You want something like this:
void vectorSum(double vecSum[], double const vec1[], double const vec2[], int size){
for(int i= 0; i %26lt; size; i++){
vecSum[i]= (vec1[i] + vec2[i]);
}
Now you can use your function like this:
int const v_size = 20;
int vec1[v_size];
int vec2[v_size];
int vec_sum[v_size];
// fill vec1 and vec2 with values here
vectorSum(vec_sum, vec1, vec2, v_size);
// Now vec_sum is the sum of vec2 and vec2
Reply:Some quick thoughts:
1. vec1 and vec2 haven't been defined properly.
2. vec1 and vec2 haven't been given any values.
3. Why are vec1 and vec2 inside the function? Shouldn't they be outside it?
Hope that helps.
Reply:Hello. You don't declare vec1 and vec2 as arrays. I'm just a tired old C programmer up past his bed time but it looks as if you declare them as doubles, that is as single variables, not arrays of variables, and certainly do not initialize them. If they are supposed to be global variables, why do you declare them in the block? And why aren't they arrays if you're going to treat them as arrays? If I were a good compiler, I'd assign a space for a double called vec1 and next to it a space for a double called vec2. Then if I were a good but nasty compiler, I would on the vecSum[i] line add vec1 and vec2 for 0, vec2 and whatever garbage is next to it for 1, the garbage I used for vec2[1] for the new vec1[2] and the garbage next to that...
More likely your compiler isn't nasty so it's just declaring an error and refusing to compile.
You could possibly declare them pointers and in C terms malloc a memory length of size*sizeof(double) for each. Actually you just shouldn't declare them in this function. You should pass them as parameters.
Sunday, August 2, 2009
Two math questions?
Q1. A cubical box is to be built so that it holds 125 cubic centimetres. How precisely should the edge be made so that the volume will be correct to within 3 centimetres? (I am pretty sure this has something to do with differentials but still not able to figure out the answer).
Q2. Let A, B and C be the points with coordinates (2,1,1) (0,4,1) and (2,1,4) respectively. Find the equation of the plane through the three points A, B and C. (vector question although i still am not able to figure out what to do with the coordinates).
Two math questions?
Just a guess..in Calc? lol...definately a differentials problem. Assuming you're in calc 3..because you're into vectors..try Linearization if you haven't already. I'd love to help you..but it's late..and i go brain dead over summer..sorry.
Reply:plzz...dont make us confused and frightened.
Q2. Let A, B and C be the points with coordinates (2,1,1) (0,4,1) and (2,1,4) respectively. Find the equation of the plane through the three points A, B and C. (vector question although i still am not able to figure out what to do with the coordinates).
Two math questions?
Just a guess..in Calc? lol...definately a differentials problem. Assuming you're in calc 3..because you're into vectors..try Linearization if you haven't already. I'd love to help you..but it's late..and i go brain dead over summer..sorry.
Reply:plzz...dont make us confused and frightened.
Does anyone know how to find the circumcentre of a triangle using vectors?
its just for any triangle with sides are a, b, c and vertices (x1,y1), (x2,y2) and (x3,y3). are there any theorems i could look up?
Does anyone know how to find the circumcentre of a triangle using vectors?
I don't know the theorem, but the circumcentre is :
( (x1+x2+x3)/3 , (y1+y2+y3)/3 ). If you treat the vertices as a vector, then the vector for the circumcentre is
1/3* ( (x1,y1) + (x2,y2) +(x3,y3) )
Reply:I am not from an English speaking country, so it is sometimes difficult to understand the meaning of some terms.
As I understand, a circumcenter is a center of a circle which comes through the three points A(x1, y1), B(x2,y2), C(x3,y3).
The previous answer gave the center of mass of a triangle with these points as apexes, (x1 + x2 + x3)/3 and (y1 + y2 + y3)/3.
The center of a circle wich comes through the points A, B, C is equally remote from the three points. Geometrically, it is the intersection point of lines that are perpendicular and come across the middle points of sections AB, BC and/or AC.
To find it, we can write and solve two equations for the lines that come across the middles of triangle's sides and perpendicular to the sides.
So, for the AB side, the middle point is {(x1+x2)/2, (y1+y2)/2}. The coefficient k for the line y=kx + a we can find from the condition that for two perpendicular to each other lines
Y = KX + A and y = kx + a, K*k = -1, and we know the coefficient for the AB side of the triangle as (y2-y1)/(x2-x1), hence, the line is
y = - (x2-x1)/(y2-y1)x + a, and to find a we have to take
x = (x1+x2)/2, and to obtain y = (y1+y2)/2:
(y1+y2)/2 = - [(x2-x1)/(y2-y1)](x1+x2)/2 + a
a = (y1+y2)/2 + (x2-x1)(x2+x1)/[2(y2-y1)].
So, the first line is:
y(AB) = - (x2-x1)/(y2-y1) x + (y1+y2)/2 + (x2-x1)(x2+x1)/(2(y2-y1))
The second line we shall write for the perpendicular to the BC side of the triangular that comes across the middle of the BC side. Making the analogous calculations, we shall obtain
y(BC)= - (x3-x2)/(y3-y2) x + (y2+y3)/2 + (x3-x2)(x3+x2)/(2(y3-y2))
And the intersection of these two lines will give us the centre of the circle. So, we only have to find the x and y common to these two equations. We have to find x, at which y(AB) = y(BC)
- (x3-x2)/(y3-y2) x + (y2+y3)/2 + (x3-x2)(x3+x2)/(2(y3-y2)) =
= - (x2-x1)/(y2-y1) x + (y1+y2)/2 + (x2-x1)(x2+x1)/(2(y2-y1))
x[(x2-x1)/(y2-y1) - (x3-x2)/(y3-y2)] = [(y1+y2) - (y3+y2)]/2 +
(x2-x1)(x2+x1)/(2(y2-y1)) - (x3-x2)(x3+x2)/(2(y3-y2)), and
x[y3(x2-x1) +y1(x3-x2) +y2(x1-x3)]/[(y2-y1)(y3-y2)] = (y1-y3)/2 +
+ (x2-x1)(x2+x1)/2(y2-y1) - (x3-x2)(x3+x2)/2(y3-y2),
or
x = (1/2)[(y2-y1)(y3-y2)(y1-y3) + (x2-x1)(x2+x1)(y3-y2) - (x3-x2)(x3+x2)(y2-y1)]/[y3(x2-x1) + y1(x3-x2) + y2(x1-x3)] =
= X = (1/2)[(y2-y1)(y3-y2)(y1-y3) + y3(x2^2-x1^2) +y1(x3^2-x2^2) +y2(x1^2-x3^2)]/[y3(x2-x1) + y1(x3-x2) + y2(x1-x3)]
This is the x value for the centre of the circle. And we can find the y value for the centre of the circle. For example, from y(AB) = - (x2-x1)/(y2-y1) x + (y1+y2)/2 + (x2-x1)(x2+x1)/(2(y2-y1)), we shall input the value of x and obtain:
Y= - 0.5 [(x2-x1)(x1-x3)(x3-x2) - y3^2(x2-x1) - y1^2(x3-x2) - y2^2(x1 - x3)]/[y3(x2 - x1) + y1(x3 - x2) + y2(x1 - x3)] =
= (1/2)[(x2-x1)(x1-x3)(x3-x2) - y3^2(x2-x1) - y1^2(x3-x2) - y2^2(x1-x3)]/[x1(y3-y2) +x2(y1-y3) + x3(y2-y1)] =
= Y = (1/2)[(x2-x1)(x1-x3)(x3-x2) + x1(y3^2-y2^2) + x2(y1^2-y3^2) + x3(y2^2-y1^2)]/[x1(y3-y2) +x2(y1-y3) + x3(y2-y1)]
So, as we see, the formula for the centre of the circle coordinates exists, and the expression is symmetrical concerning the substitution of x to y coordinates of the points, and also symmetrical concerning the substitution of (x1,y1), (x2,y2) and (x3,y3) between each other.
send flowers
Does anyone know how to find the circumcentre of a triangle using vectors?
I don't know the theorem, but the circumcentre is :
( (x1+x2+x3)/3 , (y1+y2+y3)/3 ). If you treat the vertices as a vector, then the vector for the circumcentre is
1/3* ( (x1,y1) + (x2,y2) +(x3,y3) )
Reply:I am not from an English speaking country, so it is sometimes difficult to understand the meaning of some terms.
As I understand, a circumcenter is a center of a circle which comes through the three points A(x1, y1), B(x2,y2), C(x3,y3).
The previous answer gave the center of mass of a triangle with these points as apexes, (x1 + x2 + x3)/3 and (y1 + y2 + y3)/3.
The center of a circle wich comes through the points A, B, C is equally remote from the three points. Geometrically, it is the intersection point of lines that are perpendicular and come across the middle points of sections AB, BC and/or AC.
To find it, we can write and solve two equations for the lines that come across the middles of triangle's sides and perpendicular to the sides.
So, for the AB side, the middle point is {(x1+x2)/2, (y1+y2)/2}. The coefficient k for the line y=kx + a we can find from the condition that for two perpendicular to each other lines
Y = KX + A and y = kx + a, K*k = -1, and we know the coefficient for the AB side of the triangle as (y2-y1)/(x2-x1), hence, the line is
y = - (x2-x1)/(y2-y1)x + a, and to find a we have to take
x = (x1+x2)/2, and to obtain y = (y1+y2)/2:
(y1+y2)/2 = - [(x2-x1)/(y2-y1)](x1+x2)/2 + a
a = (y1+y2)/2 + (x2-x1)(x2+x1)/[2(y2-y1)].
So, the first line is:
y(AB) = - (x2-x1)/(y2-y1) x + (y1+y2)/2 + (x2-x1)(x2+x1)/(2(y2-y1))
The second line we shall write for the perpendicular to the BC side of the triangular that comes across the middle of the BC side. Making the analogous calculations, we shall obtain
y(BC)= - (x3-x2)/(y3-y2) x + (y2+y3)/2 + (x3-x2)(x3+x2)/(2(y3-y2))
And the intersection of these two lines will give us the centre of the circle. So, we only have to find the x and y common to these two equations. We have to find x, at which y(AB) = y(BC)
- (x3-x2)/(y3-y2) x + (y2+y3)/2 + (x3-x2)(x3+x2)/(2(y3-y2)) =
= - (x2-x1)/(y2-y1) x + (y1+y2)/2 + (x2-x1)(x2+x1)/(2(y2-y1))
x[(x2-x1)/(y2-y1) - (x3-x2)/(y3-y2)] = [(y1+y2) - (y3+y2)]/2 +
(x2-x1)(x2+x1)/(2(y2-y1)) - (x3-x2)(x3+x2)/(2(y3-y2)), and
x[y3(x2-x1) +y1(x3-x2) +y2(x1-x3)]/[(y2-y1)(y3-y2)] = (y1-y3)/2 +
+ (x2-x1)(x2+x1)/2(y2-y1) - (x3-x2)(x3+x2)/2(y3-y2),
or
x = (1/2)[(y2-y1)(y3-y2)(y1-y3) + (x2-x1)(x2+x1)(y3-y2) - (x3-x2)(x3+x2)(y2-y1)]/[y3(x2-x1) + y1(x3-x2) + y2(x1-x3)] =
= X = (1/2)[(y2-y1)(y3-y2)(y1-y3) + y3(x2^2-x1^2) +y1(x3^2-x2^2) +y2(x1^2-x3^2)]/[y3(x2-x1) + y1(x3-x2) + y2(x1-x3)]
This is the x value for the centre of the circle. And we can find the y value for the centre of the circle. For example, from y(AB) = - (x2-x1)/(y2-y1) x + (y1+y2)/2 + (x2-x1)(x2+x1)/(2(y2-y1)), we shall input the value of x and obtain:
Y= - 0.5 [(x2-x1)(x1-x3)(x3-x2) - y3^2(x2-x1) - y1^2(x3-x2) - y2^2(x1 - x3)]/[y3(x2 - x1) + y1(x3 - x2) + y2(x1 - x3)] =
= (1/2)[(x2-x1)(x1-x3)(x3-x2) - y3^2(x2-x1) - y1^2(x3-x2) - y2^2(x1-x3)]/[x1(y3-y2) +x2(y1-y3) + x3(y2-y1)] =
= Y = (1/2)[(x2-x1)(x1-x3)(x3-x2) + x1(y3^2-y2^2) + x2(y1^2-y3^2) + x3(y2^2-y1^2)]/[x1(y3-y2) +x2(y1-y3) + x3(y2-y1)]
So, as we see, the formula for the centre of the circle coordinates exists, and the expression is symmetrical concerning the substitution of x to y coordinates of the points, and also symmetrical concerning the substitution of (x1,y1), (x2,y2) and (x3,y3) between each other.
send flowers
Physics (Newton's 2nd Law, Constant Acceleration, Vectors)?
A block is projected up a frictionless inclined plane with initial speed Vo = 3.42 m/s. The angle of incline is = 32.1°.
(a) How far up the plane does the block go?
_________ m
(b) How long does it take to get there?
_________ s
(c) What is its speed when it gets back to the bottom?
_________ m/s
I tried, but didn't reach the right result.
Physics (Newton's 2nd Law, Constant Acceleration, Vectors)?
Easy peasy :)
Firstly, you need to translate the vector of initial movement into its two axis, one going up and down, the same direction as gravity is going, and the other side-to-side.
This can be done using trigonometry. The speed in up/down can be determined using a triangle with hypotenuse being 3.42m/s and the incline angle of 32.1 degrees. sin(32.1) = y/3.42. Y = 1.82m/s. Then, determine the side to side componant with Pythagorean Theorem: 3.42^2 = X^2 + 1.82^2. X = 2.9m/s
Now that you have that figured out, you can determine the amount of time it takes for the object to accelerate to 0m/s (from Y=1.82m/s) using the At + Vo = V(t) formula. Given that gravity is 9.81m/s/s we have:
-9.81m/s/s * t = -1.82m/s
-9.81m/s/s * t = -1.82m/s
t = 0.185 seconds (b)
You can determine (a) by using the At^2 + Vt + Xo = X(t) formula by plugging in t = 0.185 seconds.
-9.81m/s/s * t^2 + 1.82m/s * t = X(t)
-9.81m/s/s * (0.185)^2 + 1.82m/s * 0.185 = X(t)
-0.3357m + 0.3367m = X(t) = 0.001m ( 1mm )
The speed when it comes back down can be determined by seeing how long it falls from its highest point (1mm) and then adding in a side-ways componant.
At^2 + Vt + 1mm = X(t)
At^2 + Vt = -0.001m
-9.81m/s/s * t^2 = 0.001m
t^2 = 0.0001
t = 0.01s (this means the object falls for 0.01 seconds before being at the starting position/bottom)
At = Vy... -9.81m/s/s * 0.01 = -0.1m/s
Plugging this value back into the original starting triangle, putting it at the up and down spot, we can see see that:
sin(32.1) = -0.1 / X
X = 0.188m/s (c)
Hopefully my math isnt wrong. :)
(a) How far up the plane does the block go?
_________ m
(b) How long does it take to get there?
_________ s
(c) What is its speed when it gets back to the bottom?
_________ m/s
I tried, but didn't reach the right result.
Physics (Newton's 2nd Law, Constant Acceleration, Vectors)?
Easy peasy :)
Firstly, you need to translate the vector of initial movement into its two axis, one going up and down, the same direction as gravity is going, and the other side-to-side.
This can be done using trigonometry. The speed in up/down can be determined using a triangle with hypotenuse being 3.42m/s and the incline angle of 32.1 degrees. sin(32.1) = y/3.42. Y = 1.82m/s. Then, determine the side to side componant with Pythagorean Theorem: 3.42^2 = X^2 + 1.82^2. X = 2.9m/s
Now that you have that figured out, you can determine the amount of time it takes for the object to accelerate to 0m/s (from Y=1.82m/s) using the At + Vo = V(t) formula. Given that gravity is 9.81m/s/s we have:
-9.81m/s/s * t = -1.82m/s
-9.81m/s/s * t = -1.82m/s
t = 0.185 seconds (b)
You can determine (a) by using the At^2 + Vt + Xo = X(t) formula by plugging in t = 0.185 seconds.
-9.81m/s/s * t^2 + 1.82m/s * t = X(t)
-9.81m/s/s * (0.185)^2 + 1.82m/s * 0.185 = X(t)
-0.3357m + 0.3367m = X(t) = 0.001m ( 1mm )
The speed when it comes back down can be determined by seeing how long it falls from its highest point (1mm) and then adding in a side-ways componant.
At^2 + Vt + 1mm = X(t)
At^2 + Vt = -0.001m
-9.81m/s/s * t^2 = 0.001m
t^2 = 0.0001
t = 0.01s (this means the object falls for 0.01 seconds before being at the starting position/bottom)
At = Vy... -9.81m/s/s * 0.01 = -0.1m/s
Plugging this value back into the original starting triangle, putting it at the up and down spot, we can see see that:
sin(32.1) = -0.1 / X
X = 0.188m/s (c)
Hopefully my math isnt wrong. :)
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