What is the integral of rootsquare 1 plus x2 over x without using partial fraction?
Int sqrt(1+x2)/x = sqrt(1+x2) + LN [(sqrt(1+x2) - x -1) / (sqrt(1+x2) - x +1)]
A quadrilateral is a parallelogram if a pair of opposite sides is equal and parallel prove?
A quadrilateral is a parallelogram if one pair of opposite sides are equal and parallel
Let ABCD be a quadrilateral in which ABCD and AB=CD, where means parallel to.
Construct line AC and create triangles ABC and ADC.
Now, in triangles ABC and ADC,
AB=CD (given)
AC = AC (common side)
Angle BAC=Angle ACD
(corresponding parts of corresponding triangles or CPCTC)
Triangle ABC is congruent to triangle CDA by Side Angle Side
Angle BCA =Angle DAC by CPCTC
And since these are alternate angles, ADBC.
Thus in the quadrilateral ABCD, ABCD and ADBC.
We conclude ABCD is a parallelogram. var content_characters_counter = '1032';
A square. All squares are parallelograms, but not all parallelograms are squares.
What are the numbers from 1 to 200 have an expanded form that is identical to the standard form?
1,2,3,4,5,6,7,8,9,10,20,30,40,50,60,70,80,90,100,200
How fast would a 75 kilo male fall from 12000 feet?
The acceleration of an object that falls from a certain height does not depend on its mass, in an ideal condition with no air resistance.
The value of acceleration is the acceleration due to gravity, which is 9.81 m s-2.
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However, in this case, air resistance is going to matter. 12000 feet is high enough for the person to accelerate to what we call terminal velocity. Terminal velocity is the velocity where the force of acceleration due to gravity (9.81 m s-2) is matched by the air resistance. That velocity varies, depending on the outline shape of the person, and is typically around 200 km/h or 125 mph. That will be the velocity of the fall.
How do you show that a square matrix A is similar to its transpose?
First we will handle the diagonalizable case.
Assume A is diagonalizable, A=VDV-1.
Thus AT=(V-1)TDVT,
and D= VT AT(V-1)T.
Finally we have that A= VVT AT(V-1)TV-1, hence A is similar to AT
with matrix VVT.
If A is not diagonalizable, then we must consider its Jordan canonical form,
A=VJV-1, where J is block diagonal with Jordan blocks along the diagonal.
Recall that a Jordan block of size m with eigenvalue at L is a mxm matrix having L along the diagonal and ones along the superdiagonal.
A Jordan block is similar to its transpose via the permutation that has ones along the antidiagonal, and zeros elsewhere.
With this in mind we proceed as in the diagonalizable case,
AT=(V-1)TJTVT.
There exists a block diagonal permutation matrix P such that
JT=PJPT, thus J=PTVT AT(V-1)TP.
Finally we have that A= VPTVT AT(V-1)TPV-1, hence A is similar to AT
with matrix VPTVT.
Q.E.D.
What does warranted 12 dwt mean?
It means you have a piece of silverplate that has 12 pennyweight of silver plated onto a base metal.
2*(L*B + B*H + H*L) cubic units where
L = length
B = Breadth
H = Height.
What is the geographic grid used on the globe is based on the division of a circle in what degrees?
360 degrees from pole to pole and 180 degrees in circles parallel to the equator.
What is a true about the AAA theorem and the SSS postulate?
There is nothing true about the AAA theorem and the SSS postulate because the AAA postulate is not true!
How do you prove a conjecture is false?
Prove that if it were true then there must be a contradiction.
What number that is fewer than 40 when divided in group of 5 leave a balance of 3?
3, 8, 13, 18, 23, 28, 33, and 38 are all solutions.
Show that convergent sequence is bounded?
Wrong answer above. A limit is not the same thing as a limit point. A limit of a sequence is a limit point but not vice versa. Every bounded sequence does have at least one limit point. This is one of the versions of the Bolzano-Weierstrass theorem for sequences. The sequence {(-1)^n} actually has two limit points, -1 and 1, but no limit.
Let M be the subset of 2x2 matrices A such that det(A)=0 and tr(A'A)=4 (I shall use ' to denote transpose).
Recall that one of the three definitions of a k-dimensional manifold is:
M c R^n is a k-dimensional manifold if for any p in M there is a neighborhood W c R^n of p and a smooth function F:W-->R^n-k so that F^-1(0) = M intersection W and rank(DF(x))=n-k for every x in M intersection W.
In short, what we need to do is find a function F so that the inverse image of the zero vector under F gives M and the rank of the derivative of F is equal to the dimension of the codomain of F.
Let an arbitrary 2x2 matrix be written as:
[a c]
[b d]
Then the two constraints that define M are
1) ad-bc=0
2) a^2+b^2+c^2+d^2=norm(a,b,c,d)^2=4
Define F:R^4-->R^2 by F(a, b, c, d)=(ad-bc, norm(a,b,c,d)^2-4). Then clearly F^-1(0,0)=M. Furthermore, F is smooth on M because the multiplication and addition of smooth functions (a, b, c, d) is also smooth. (Note that we have taken the neighborhood W to be some superset of M. This guaranteed to exist because if we interpret the set of 2x2 matrices as R^4, every point in M has norm 2, so any ball centered at the origin with length greater than 2 will contain M).
All that remains to be done is to check that rank(DF(x))=2 for every x in M. Observe that [DF]= [d -c -b a]
[2a 2b 2c 2d]
We shall now argue by contradiction. Suppose rank(DF) did not equal 2 for every x in M. Then we know that the two rows are linearly dependent i.e.
h[d -c -b a] + k[2a 2b 2c 2d] = 0 and h and k are not both 0.
Suppose h is 0. Then we have 2ka = 2kb = 2kc = 2kd = 0, and since k cannot also be 0, this implies that a=b=c=d=0, therefore norm(a,b,c,d)^2=0. But, (a,b,c,d) must be in M, so this is a contradiction. Hence h cannot be 0.
Now suppose h is nonzero. Then we can divide it out and there exists a, b, c, d and a constant k so that
[d -c -b a] + k[2a 2b 2c 2d] = 0 i.e. we have:
d + 2ka = -c + 2kb = -b + 2kc = a + 2kd = 0. From this we can substitute to obtain:
a(1 - 4k^2) = b(4k^2 - 1) = c(4k^2 - 1) = d(1 - 4k^2) = 0 and hence
a^2(1 - 4k^2)^2 = b^2(4k^2 - 1)^2 = c^2(4k^2 - 1)^2 = d^2(1 - 4k^2)^2 = 0.
Note that (1 - 4k^2)^2 = (4k^2 - 1)^2. Now, adding the four above expressions together we get:
(a^2 + b^2 + c^2 + d^2)(1 - 4k^2)^2 = norm(a,b,c,d)^2(1 - 4k^2)^2 = 0. But, since we require (a,b,c,d) to be in M, this reduces to 4(1 - 4k^2)^2 = 0. This implies that
1 - 4k^2 = 0, and hence k = +/-(1/2).
Now, if k=+1/2, then we have a + d = b - c = 0, therefore d = -a and b = c. Since we have det(A) = 0 as one of our constraints on M, this implies that ad - bc =
-(a^2) - (b^2) = -(a^2 + b^2) = 0, which implies a = b = 0, by the property of norms. But, if a = b = 0, then (a,b,c,d) = (0,0,0,0) and hence norm(a,b,c,d)^2 = 0, which is a contradiction.
We can argue analogously for the case where k = -1/2. Hence, assuming that rank(DF) is not 2 for some (a,b,c,d) in M leads to a contradiction, so we conclude that rank(DF)=2 for all x in M.
Finally, from this result, we conclude that since F:R^4-->R^2=R^(4-2), M must be a 2-dimensional manifold.
It very much depends on f.
If f is one-to-one and onto (injective and surjective) then yes, else no.
One-to-one means that for each element in the domain there is a different image in the range. This is not true for g(x) = x2 for example, where -3 and +3 are both mapped to +9. So g(x) does not have an inverse UNLESS you restrict the domain of g to non-negative reals. Then -3 is no longer in the domain.
Onto means that every element in the range of the function has a corresponding element in the domain which is mapped onto it. Again, a suitable changes to the domain and range can transform a function without an inverse into an invertible one.
Prove that intersection of two convex polgons is a convex as well?
I will prove a more general theorem from which your answer follows immediately. Theorem: The intersection of any number (including 2) of convex polygons is convex.
Proof
Let C be the intersection of Ci which is a set of iconvex polygons. By definition of intersection, if two points A and B belong to C then they belong to every one of the Ci . But the convexity of each of the Ci tells us that line segment AB is contained in Ci . Therefore, the line segment AB is in C and because ABwas arbitrary we conclude that C is convex
D = diagonal, S = side, P = perimeter
S = D/sqrt(2)
P = 4S = 4D/sqrt(2)
So, for this problem:
P = 48/sqrt(2)
To the nearest tenth, this is 33.9 inches