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12 sounds good for me.
You are working in 3D. So the coordinate system has three axes.
Because 2×2×2 = 8 you have 8 vertices in V(G).
000,001,011,010,100,110,101,111
Using the definition of the edge, for example 000, 111 does not represent one because thre are 3 different numbers between them not 1 as required but 000 , 001 is an edge.
There are a few others, try to find them.
Because 1/2 = sin30 and √3/2=cos30 you can write down the expression like this:
2((sin30/sin10)-(cos30/cos10))
To be able to subtract two fraction you need same denominator so:
2((sin30×cos10-cos30×sin10)/(sin10×cos10)) same as
2(sin(30-10)/(sin20/2)) = 4sin20/sin20 = 4
I have used two trig formulae:
sina×cosb - sinb×cosa=sin(a-b) and sina×cosa = sin(2a)/2 same as 2×sina×cosa = sin(2a)
Q1:
1. Start with the secant.
First point has the coordinates (-1,1) and the second one (2,4).
2. Write down the ecuation of the secant.
For example (y-1)/(4-1)=(x+1)/(2+1) same as y-x-2=0
3. Find the gradient of the secant
m = 1
4. The tangent you are looking for is parallel with the secant so they have the same gradient m = 1
5. You have now the gradient, and you can solve an equation to find the point you need.
y = x^2 gives you y' = 2x
2x = 1 (=m)
x=1/2
Answer x=1/2
Check my calculations anyway. I can be wrong, but this is the idea.
Q2: if you make a notation /x-3/ = m (> 0) you get an equation m^2-4m-12=0.
This equation has 2 solutions m=6 and m=-2.
-2 is not positive so m=6 and x-3 = 6 or x-3 = -6.
Solutions x = 9 or x = -3.
Again, check my calculations.
First piece of advice: try not to use "a" for the root if you have an "a" as coefficient in your equation. It is confusing.
Second piece of advice: The problem can be solved using the sum and the product of the roots. The idea is that the relationship you use must be simmetrical in x1 and x2 (the roots).
So we know that x1+x2 = -b/a and x1x2=c/a
x1-x2 is not simmetrical in x1 and x2 but (and here is the trick) (x1-x2)² is.
(x1-x2)² =x1² +x2² -2x1x2 = (x1+x2)² -4x1x2= (-b/a)² -4(c/a)= 1
or b² /a² -4c/a-1=0 or b² - 4ac-a² =0 QED
I hope it will help you understand more about the sum and product of the roots.
I mean square brackets.
Actually you can simplify the solution. Instead of trying to find x and y, you can look for only one variable z.
The two vectors are linearly dependent if there is a real z so (a b) +z(c d)=0 . That means a+zc=0 and b+zd=0 or a/c=b/d or ad-bc=0
I use a square pharantesis. But I have seen many other symbols.
We actually have two problems here, because of if and only if.
First you need to prove that if z1, z2, and z3 lay on the same line then that fraction is a real number. Secondly you have to prove that if the relation is right then A,B and C are collinear points.
Use coordinate geometry after you alocate for z1 a point, let's say A(x1,y1), for z2 a point B(x2,y2) and for z3 a point C(x3,y3).You will discover, after you multiply the fraction with the conjugate expression, that the numerator is nothing else then a necessary and sufficient condition for collinear points.
x is a real number probably
If the cylinder has a radius of x then its height is 2√ (1-x ²). You can check this drawing a right-angled triangle having the hypotenuse the diagonal of the cyclic rectangle with dimensions x and height. So the volume would be V = 2πx²√ (1-x² ). You know what to do next.