Linear Algebra

Given the initial velocity V, and the angle from the ground A, the total distance travelled X will be: X = 2 V2 cos(A) sin(A) / g

where "g" is the acceleration due to gravity, on earth g is approximately 9.81 m/s2.

You will notice that the mass of the object does not affect the distance traveled. We can derive this by first determining how long the projectile will be in the air. If the initial velocity is V, then the initial vertical velocity is Vsin(A). The vertical velocity will decrease at a rate of 'g' until the vertical velocity reaches zero (known as apogee), and the projectile starts falling down. The time from launch to apogee will be Vsin(A)/g.

The time for the projectile to go up is the same as for the projectile to fall down again, so the total time in the airis 2Vsin(A)/g.

Assuming we neglect friction, the horizontal velocity is Vcos(A) and does not change. The total distance traveled horizontally is the horizontal speed multiplied by the time spend in the air. So X = 2Vsin(A)/g * Vcos(A) = 2V2cos(A)sin(A)/g.

The maximum distance is achived with an angle of 45o. The distance travelled is symmetric around this value, i.e. an angle of 50o will give the same distance as 40o, and an angle of 15owill give the same distance as 75o.

You know when an equation has a unique solution when there is only one variable in it. (APOLOGIES)

(RESPONSE: the question was categorized under "Linear Algebra". x^2 is non-linear and is thus not allowed, nor are sin x, x^3, log x, 2^x, etc etc. However, you are correct if you consider non-linear equations. Unfortunately, I am not sure there is a method to determine the number of solutions to non-linear equation.)

If there are more than one variable, each variable over the first will be free, and give you infinite solutions - with each additional variable adding another dimension to your solution.

(RESPONSE: See above response with regards to this topic being categorized under "Linear Algebra". My statement is true in Linear Algebra. Furthermore, Row Reduced Echelon Form and augmented matrices are the most fundamental concepts in Linear Algebra. Under normal circumstances, I would agree with you. However, this question was categorized under "Linear Algebra", so I presumed that the person asking the question is a college student.)

In general, you know that a system of equations has a unique solution when the row reduced echelon form of the augmented matrix has a pivot position in every column, except for the right most column which is the solution. If you do not have an augmented matrix, then the RREF will have a pivot position in every column.

Ok, linear algebra isn't my strongest area but I'll have a go (please note that all vectors are in bold, if a mathematical lower case letter is not in bold, assume it to be a scalar unless otherwise indicated).

First, I'm going to assume that you already know how to prove that the kernel and image of a linear operator, say f:U->V where U and V are vector spaces, is a vector subspace of U as this needs to be proved before going on to show that the sum of ker(f) and Im(f) are a vector space.

Now, a vector space must follow the following axioms (here A represents the addition axioms, and M the multiplication axioms):

A1) associativity u+(v+w)=(u+v)+w

A2) commutativity u+v=v+u

A3) identity there exists an element 0 in U such that v+0=v

A4) existence of an inverse for all u in U there exists an element -u in U such that u+(-u)=0

M1) scalar multiplication with respect to vector addition a(u+v)=au+av

M2) scalar multiplication with respect to field addition (a+b)u=au+bu

M3) compatibility of scalar and field multiplication (ab)u=a(bu)

M4) identity 1u=u where 1 is the multiplicative identity

ker(f)={u in U | f(u)=0} (or more strictly, the set of vectors u in U such that Au=0 i.e. ker(f) is the null space of the matrix A), Im(f)={v in V | f(u)=v for some u in U}

Let ker(f)+Im(f) = W. Show W is a vector space i.e. show W satisfies the above axioms.

A1, A2, M1, M2, and M3 are all trivial/simple to prove.

A3: since Im(f) is the set off all v in V obtained from f(u), assume Im(f) = V and as V is a vector space it must have, by definition, an additive identity 0, therefore W also contains 0.

A4: since ker(f) is the null space of the matrix A, f(0)=0. Now assume that f(u)=0 and f(-u)=v, since U is a vector space u+(-u)=0 so f(u+(-u))=f(0)=0 but f(u+(-u))=f(u)+f(-u)=0+vtherefore, 0=0+v which implies v=0, so if u goes to 0 in V, its inverse -u also goes to 0. A similar process for the image shows that if u goes to v then -u goes to -v. (Thanks to Jokes Free4Me for the help with this axiom)

M4: similar to A3 except V, again by definition, must have the multiplicative identity 1 and so also exists in W.

All the axioms are satisfied, therefore W=ker(f)+Im(f) is a vector space. Q.E.D.

7/8 + 5/6

Find the least common denominator

8- 8, 16, 24...

6 - 6, 12, 18, 24

24 seems to be the least common denominator

We know that 8x3 = 24 so 7x3=21 and 6x4=24 so 5x4=20 (whatever we do to the denominator, we do to the numerator)

Now the fractions are:

21/24 + 20/24

Add the numerators and keep the denominator of 24 the same.

(21+20)/24 = 41/24 = 1 17/24

Final answer is 1 17/24