Because selection bias. The problems that can be reduced to linear algebra make up a sizable portion of the problems that can be solved at all.

Linear algebra is generally a well-understood and simple domain. Whenever a problem can be reduced to linear algebra, however circuitously and cleverly, it usually is reduced to linear algebra and then, often times, it can be solved.

Of course, many problems can’t be reduced to linear algebra. Sometimes such problems can be solved in other ways, but many such problems are unsolved.

The end result is that if you survey all the problems we’ve ever solved along with their solutions, you see a lot of linear algebra.

You see a lot of linear transformations, linear approximations, eigenvalues and eigenvectors, inner products, change of basis, and did I mention eigenvalues? No? So eigenvalues. And eigenvectors. Lots of them.

You see differentials, and cohomology, and linear representations, and Hecke operators, and vector bundles, and many other sophisticated machinery which can be regarded as deep ways of linearizing.

Because at the end of the day, as Ehud de Shalit once told me, the only equations we can handle really well are linear equations. So we seek them, and if they don’t seem to be around, we conjure them.

A significant amount of modern math can be understood as a desperate attempt to linearize that which seems non-linear. We frequently fail, but we also succeed more often than might seem reasonable at first glance. The methods are ingenious, and the results are deep and surprising.

(a) Since the argument of cos(u) is u(x) = 1/x, you can invert each value for x to obtain u, and pass that value into the function cos(u); i.e., u = nπ + π/2, 2nπ, and (2n+1)π.

• cos(u), u = nπ + π/2

This takes the cosine of the quantity obtained by adding some integer multiple of π radians to π/2. Due to the periodicity, we can disregard higher multiples so that it boils down to taking the cosine at +/- π/2, outputting a value of zero for any integer value n. Therefore, cos(u)=0.

• cos(u), u = 2nπ

This expression takes the cosine of somepositive integer multiple of 2π. Again, take advantage of the periodicity of the cosine function, allowing us to equate as: cos(2nπ) = cos(0n) = 1 for any integer value n. Therefore, cos(u) = 1.

• cos(u), u = (2n+1)π

For values of n = {0, 1, 2, …, n}, u will be equal to odd multiples of π. Once again due to the periodicity of the cosine function, it’s the same as taking cos(π) = -1. Therefore, cos(u) = -1.

(b) In the first expression taking the limit of cos(1/x) as x approaches 0 is exactly equivalent to taking the limit of cos(u) as u approaches infinity. Since cosine is an oscillatory function, it will shift continuously between values of 1 and -1 forever as the argument goes to infinity.

As for the second expression, you can simplify it to the limit of f(x) = 1 as x approaches 0. Since x is no longer an argument in the simplified form, the limit of f(x) as x approaches 0 is 1.

(c) If we define g(x) such that f(x) = sin (x), then there will be no discontinuity at x = 0, since sin(0) = 0.

The first part is pretty self explanatory where you plug in the value of the given x into f(x). (Hint: The answer to the x=1/(nπ+π/2) is zero.)

I’m going to skip answering the second part as well as it really is nothing more than a basic limit question and i hope you’re familar with calculating limits if you’re attempting this assignment/sheet/question bank.

The only interesting part of this question is the third one, although pretty basic and easy to answer, but still comparitively interesting.

g(x)=xcos(1/x)

For any function to be continuous, it’s Left hand limit and Right hand limit at any given point should equal to the value of that function at that point. Here, we need to check the LHL and RHL at 0.

When x approaches zero, the value of Lim xcos(1/x) is zero.

How? You see, when x tends to zero the value of cosine approaches infinity, but since cosine can only give values between -1 and 1 it must be some value between these two numbers. Therefore, putting values in g(x) we get,

0×(some finite value between -1 and 1) so we get 0. (The RHL and LHL will both be equal to zero, simply replace x by (0-h) in g(x) and make h approach zero, and (0+h) in g(x) and make h approach zero to get LHL and RHL respectively)

Now,

LHL=RHL=g(0) for a continous function.

Therefore we should define g(0)=0 for it to be continuous.

You have two possible approaches. The first and most obvious is to solve for x and then substitute it into 1/3 - 7x.

You may have thought of that, but are feeling uncomfortable about the fractions. They do get really messy and that will be especially bad if this is in the calculator inactive part. So instead, solve for 1/3 -7x. Basically you need to take the original equation and manipulate it with algebra so that on one side you have 1/3 -7x and on the other side you have a number.

Consider 1/15 - 7x/5. That 7x that we want is already there. We just don't want the 5 in the denominator. So what happens if we multiply both sides by 5?

5(1/15) = 1/3 (Awesome! That's what we wanted!)

So 1/3 - 7x = 5(21) = 105.

I don’t agree with your premises, unless we use a very broad definition of both “linear algebra” and “reduce”.

Let’s start with “linear algebra”. Do we count Hilbert spaces as linear algebra? On one hand, if you think about them in an abstract way, then probably yes. But most of useful Hilbert spaces are functional spaces and their interesting properties are closely related to analytic properties. In this sense, it is more like analysis than algebra. For me, as soon as you have inequalities and estimates it’s no longer algebra, it’s analysis. So from where I stand, it looks more like analysis that use some language and notion of “linear algebra” than “linear algebra” itself.

Similar thing happens with “reduce”. If we are talking about “use”, then it is true. But, on the other hand, in most papers you have to add or multiply at least two numbers. Would you say that everything is reduced to arithmetic? Probably not. Let’s look at more advanced example: linearisation. Is it reduction to algebra? Yes and No. Again, from my experience, the hard and interesting part is how to linearize and how to show that your linearization still captures important properties of the original object. Doing a bit of linear algebra at the end is usually is almost in the same category as “reduction to arithmetic”.

To summarise: With a wide definition of “linear algebra” and “reduce” it is true since linear algebra is one of the simplest fundamental blocks of mathematics, so it appears all over the place. With more strict definition, I do not believe that it actually true.

I was thinking of writing a solution for an analogous problem, since I am not much into solving homework problems for others. But I see that someone did it for you already, so… given that you described this as an algebra problem, let me show you how to find a general solution, for any width of paper, any number of columns, any margins, etc.

Let the width of the paper be denoted by the letter [math]w[/math], and the margins by the letter [math]m[/math]. Since there are two margins, we need to subtract the width of the margin twice to get the usable width: it will be [math]w - 2m[/math].

Let the number of columns be denoted by the letter [math]n[/math], and the width of each column by the letter [math]c[/math]. Then the total width occupied by [math]n[/math] columns (without gutters) would be [math]nc[/math].

But in between the [math]n[/math] columns, there are also [math]n-1[/math] gutters. Let the ratio of the width of a gutter to the width of a column be denoted by the letter [math]g[/math]. Then the width of one gutter will be [math]gc[/math], and the width of all the [math]n-1[/math] gutters together will be [math](n-1)gc[/math].

Together, the widths of the columns and the widths of the gutters must be the same as the total width minus the two margins. This gives us a nice equation:

[math]nc + (n-1)gc = w - 2m.[/math]

This equation can be solved for any of the symbols it contains, but it so happens that in your exercise, all the symbols have assigned values except for one: [math]c[/math], the width of a column. So we need to solve this equation for [math]c[/math], which is very easy to do, since c is a common factor on the left-hand side. First, I rewrite the equation to show this:

[math][n + (n-1)g]c = w-2m,[/math]

next, I divide by the factor that multiples c on the left-hand side:

[math]c = \dfrac{w-2m}{n + (n-1)g}.[/math]

Go ahead, try it: plug in the values for [math]w = 8.5''[/math], [math]m = 1''[/math], [math]n = 3[/math], [math]g = 1/8[/math]. See what you get.

The beauty of doing things this way is that you can now use the formula for other combinations of numbers. A good example is [math]w = 8.5''[/math], [math]m = 0.5''[/math], [math]n = 4[/math], [math]g = 1/3[/math]. That was the analogous problem that I wanted to use in my answer before I realized that someone already offered a solution of the original question.

I have the same answer as with Andre Rajoo, but since it was not in the choices, I tried to find the solution that would fit. And I found a solution that fits. First if we compare their rate for the first 2 years (2870/2) with their 5 year rate (7150/5) we could see that they are different. 1435 per year compared to 1430 per year. With this we could assume that their yearly rate of savings differed for the last 3years compared to the first 2. So to find the current yearly rate i subtract 2870 from 7150 and divide the result by 3. I have got the current yearly rate of 1426.67 per year. I multiplied that by 1.5 and add that to their current savings and I arrived with 9290 which is option C.

I mean no disrespect to the other people who have answered, but this question assumes around a 10th grade mathematics background. It is designed to be answered in less than 1 minute. It certainly does not use trig.

You want to maximize 1/2 BH. Just eliminate answers.

If the answer was E then we would need a triangle with height 28 and base 10. It should be obvious this can't work if you draw the picture.

If it were D then you would need a triangle with height 14 and base 10. This obviously will not work either.

Why could we eliminate the answers? No matter what, to calculate the height we'd have to use 7 as a hypotenuse of a right triangle. You can't have legs of that right triangle be longer.

This takes about 20 seconds to solve if you're really over thinking it.

Edit:

This answer seems to have angered some people who don’t understand it. I’ll try to explain it better. I suggest you draw a picture to understand.

The question requires us to have two sides be of length 10 and 7. For the purpose of this answer I’ll choose the base to be 10. It will work if the base is 7 as well.

If the base is 10 in case E we know we have a height of 28. This height subdivides the triangle into two right triangles. The hypotenuses of these right triangles are sides of our big triangle. We know one of them MUST be 7. Therefore, one of these triangles has a hypotenuse of 7 and a leg of 28. This is a CONTRADICTION. We have a leg of a triangle that is bigger than the hypotenuse.We know this situation must be impossible. The height is forced to be a leg and 7 is forced to be the hypotenuse. If the height is bigger than 7 then this triangle physically cannot exist. This allows us to eliminate answers E and D. You’ll find you cannot eliminate answer C.

The only mathematics this requires is the Pythagorean theorem, or simply the knowledge that if a^2+b^2=c^2 and we’re dealing with real numbers then a<c and b<c. Pretty much common sense. Up until 2016, the SAT did not require trigonometry at all. This answer was written in 2015. So if your answer uses trig then you’re making it unnecessarily difficult.

1. For the definitions, rational numbers are any numbers you can represent as a fraction with integers in the numerator and denominator, except for zero in the denominator.
2. Irrational numbers are those “between” the integers that are not rational. These would be numbers whose decimal representations go on infinitely without a repeating pattern. These would be numbers like pi, square roots that can’t be reduced to integers. In the list, the only irrational numbers are sqrt 2, sqrt 26, sqrt 5, pi/2, and 2pi.
3. To place the numbers on the number line, just estimate. Use 3.14 for pi to figure the ones with pi. For sqrt 26, think: sqrt 25 =5, sqrt 36 =6, so sqrt 26 would be between 5 and 6.

Option B

Others have already answered it.
These types of questions are the form of non-terminating recurring decimal notation.
See 24 is recurring after decimal point.

These types of questions are solved by multiplyting the given number repeatedly with 10 until you get the same recurring number after decimal point and then subtract with its counterpart having same recurring term after decimal.

1. Multiply the number with 10, you get 12.42424....
2.Multiply again with 10, you get 124.24242424....

The recurring term for 100x is same as x, so subtract x from 100x, you get

99x = 123
=> x = 123/99 = 41/33 = p/q

=> p = 41 and q = 33

Therefore, p+q = 74.

Hence, Option B is correct.

The key to solving problems is to read carefully. If you have a textbook you can write in, then learn to underline the key information. If you are doing it online, use your highlighter. So, read carefully and determine exactly what the question is that you must answer. Write it down.

Then, write down any formulas you know which would apply to the question. For example, if it is a question of finding the length of a rectangle given the area, write down the formula for the area of a rectangle. If you don’t know it, then this may be what is causing you problems: failure to memorize important formulas.

Then, write down (or underline) the information in the problem that you know which applies to the question. In my example, you would write down the area formula for the rectangle.

Then select variables to represent the information you are looking for. Use those variables to express any other unknown information. For example, to find the area of a rectangle, you need length and width. The length is unknown, so you have already identified a variable to represent it. Is there any information in the problem about the width? It may tell you the width, or it may give you information about the width in terms of the length (“the width is 2 cm less than the length”). You need to express that as a mathematical expression. (“if x= length, then width is x-2”).

Now fill these values in the formula. There you have your equation. Now, solve it.

Is your question “how do I solve an equation?” In that case, it depends on the type of equation. If it is linear, you must solve for the variable you are trying to find (get that variable by itself on one side of the equation). If it is quadratic, you must get 0 on one side, then factor and use the zero product property (if two numbers are multiplied together and the result is 0, then one of them must be zero). I doubt it is any higher than that, or you wouldn’t be asking the question you are asking.

The content is reproduced, as text, below.

A plane is flying at a speed of 300 m/s and maintaining an altitude of 10000 m when a bolt becomes detached. Ignoring air resistance, find

(i) the time that the bolt takes to reach the ground

(ii) the horizontal distance between the point where the bolt leaves the plane and the point where it hits the ground

(iii) the speed of the bolt when it hits the ground

(iv) the angle to the horizontal at which the bolt hits the ground.

As others have suggested, the usual approach would be to consider independent hoizontal and vertical quantities. The horizontal velocity of 300 m/s would not affect the vertical motion.

(i) This answer is then the time of freefall from 10000 m.

By blind memorisation or otherwise, typically here we would use an equation like

[math]y=u_yt+\frac{1}{2}a_yt^2[/math]

. There is no initial vertical velocity ([math]u_y=0[/math]) and the vertical acceleration is the effect of gravity so ultimately

[math]t=\sqrt{\frac{2y}{g}}[/math]

and we could shove the quantities in for the calculation.

(ii) There is no cause for the horizontal velocity to change, so for an object travelling at the given horizontal velocity over this time above,

x=v_xt=v_x\sqrt{\frac{2y}{a_y}}

. Again, substitution is straightforward.

(iii) Just to introduce another approach, we consider conservation of energy. We are interested in kinetic energy ([math]K=\frac{1}{2}mv^2[/math]) and potential energy ([math]U=mgh[/math]) amounts.

At the top, the object has no vertical kinetic energy ([math]K_y=0[/math]) but all the gravitational potential energy ([math]U_y=mgy[/math]) of being at its initial altitude. At the bottom, it has no gravitational potential energy relative to ground ([math]U_y=0[/math]) but plenty of vertical kinetic energy ([math]K_y=\frac{1}{2}mv_y^2[/math]).

Conservation of energy means total energy at the top must be equal to total energy at the bottom.

[math]0+mgy=0+\frac{1}{2}mv_y^2[/math]

We can rearrange to make [math]v_y[/math] the subject.

[math]v_y=\sqrt{2gy}[/math]

Overall speed is the length of the overall velocity vector, which has the above vertical component, and a horizontal component given in the question. How to find this length has already been covered elsewhere.

(iv) To find angle above horizontal (call it [math]A[/math]), note that trigonometric tangent is given by the vertical component divided by the horizontal component.

[math]tanA=\frac{v_y}{v_x}[/math]

With all the rearrangements up to this point, finding that angle should again be straightforward enough.

[math]A = Arctan(\frac{\sqrt{2gy}}{v_x})[/math]

I would suggest starting from linear algebra.

Matrices, operations with matrices, their determinants.

Different methods for solving linear equation sets: general direct methods, iterative methods, solving equation sets with sparse matrices, with very big matrices; pre-conditioning, regularization. Methods for symmetric matrices. Cholesky factorization of a matrix. Problems of eigenvalues and eigenvectors and different algorithms of their computing. Higher algebra (rings, groups, etc.) can be probably skipped nd postponed for a later time. If you have any specific questions, related to specific problems - you are welcome

You need to use the concept of relative velocity to solve this problem. But before that I will take the liberty to add few details to the question.

Let's say the difference between point A and point B is D. And for this journey to be feasible the bird should fly with a speed v > 10kmph .

Scenario A. Point A is behind point B

Bird has started it's journey to point B at time t = 0. It reaches the point B at time t1. Which is given by the formula

Time = distance / speed

Now when bird is flying from point A to point B with a speed of v. The B is also moving ahead of the bird with a speed of 10 kmph. So when bird covers v km in an hour. Train has moved forward by 10 km. So the distance to point B has been reduced to V - 10 km. In other words velocity of bird relative to train while the bird flies towards point B is vr1 = v - 10

So t1 = D / ( v - 10 )

When bird is flying back to point A. The train is also moving towards the bird with a speed of 10kmph. So vr2, that is the velocity of bird relative to point A is v + 10
Therefore t2 = D / ( v + 10 )

So total time t = t1 + t2 = 2Dv/ ( v^2 - 100 )

Scenario 2: point A is ahead of point B

The scenario is are reversed but you get the same answer
t1 = D / ( v + 10 )
t2 = D/(v-10)
t = 2Dv/(v^2 - 100 )

Substitute the values from the book for v and D and you get the answer. Please let me know if you need any more clarification on my answer

Note: if the question is about the person walking inside the train instead of bird flying over it. The answer will be entirely different.

There will always be questions that have no “simple” solution and your definition of simple will always depend on your level of skill, knowledge, creativity and interrest. And this sentence alone says it all, but i`ll explain it to you anyways to get a little bit more into the detail of what i intended to say:

there is no universal solution for something as complex and extensive as mathematics but by training you can dive into the topic and then you`ll start to get better and better and in the end a topic which when you first came across it was extremely difficult to you appears to be supprisingly easy.

Can only agree with the other guys:

Let Vv,Vh,Dh,Dv be the vertical and horizontal speed and horizontal and vertical distance.(depending on the time t by the way)
I’ll

So the formulas are
Dv(t)=10000–0.5*g*t^2
Vv(t)=-g*t

Dh(t)=300*t
Vh(t)=300

Now to the questions:
(i) you need to find t so that Dv(t)=0
(the bolt lands on the ground which is lokated at Dv=0) Horizontal distance is unimportant for this part.)

(ii) take the t you found in (i) and put it in Dh(t). The result is the horizontal distance that was asked.

(iii) this one is trickier since you need to consider both horizontal and vertical speed.

If we are correctly, speed is actually a vector with the 2 components Vv(t) and Vh(t).

as you may know the absolute value of the speed vector V can be found as
V(t)=sqt(Vh(t)^2+Vv(t)^2)

Put the same t as before in this equation.
The result is the asnwer for this question

(iv) Many ways to solve this but the ost common one is probably using the scalar product of the V(t) vector and the horizontal vector which is parallel to the x-axis ( Vector e1=(1,0))

So we have V(t)*e1=|V(t)|*|e1|*cos(alpha)

alpha is the angle you search.

Again with the t from (i), we have given that

V(t)=(Vh(t),Vv(t)), |V(t)| =the result from (iv),
|e1|=1 as you can easily calculate, e1=(1,0).

So finding that angle shouldn’t be too hard.

To your wrong result. You should check again if you made any mistakes anywhere but it is also possible that the book’s solution is wrong.

Unfortunately this happens more often than it should, creating confusion every time :-(

A general tip for this kind of problems:

You should first write down the formulas for acceleration, speed and distance.

Also consider that speed is the integral of the accelration and the same applies to speed-distance.
I usually don’t remember the ugly distance formula.

But rather do the following:
Look what’s constant.
In your case the horizontal speed is constant. Therefor write Vh(t)=c.
And integrate that to get the distance function .

For vertical stuff, you have to go down a level deeper and start with Accv(t)=constant.
Integrate to get vertical speed and again to get vertical distance.

What the constant thing actually is, depends on how you chose your coordinate system.

I chose it’s origin at the point on the ground directly below the plane.

And never forget : Horizontal and vertical things happen independently from each other.
That’s why looking at the vertical distance was enough to find the time in (i).

First let's try to develop a general theory for these kinds of questions. So lets consider a general equation of Conic section which is

[math]ax^2+2hxy+by^2+2fx+2gy+c = 0[/math]

Rearrange this equation as

[math]ax^2+2x(hy+f)+(by^2+2gy+c) = 0 [/math]

Solving for the [math] x[/math] we have

[math]x= \dfrac{-2(hy+f)\pm2\sqrt{(hy+f)^2-a(by^2+2gy+c)}}{2a} [/math]

Now if we want that given equation should have linear factors then this also means that above expression should have linear relationship between [math]x[/math] and [math]y[/math]. Now for this to happen the term under square-root should be a perfect square.

[math]\implies (hy+f)^2-a(by^2+2gy+c)[/math] should be perfect square

Simplifying this equation and we will get

[math](h^2-ab)y^2+2y(gh-af)+(g^2-ac) = 0 [/math]

Now this equation will be a perfect square if it has only one root i.e it's discriminant is zero.

[math]\implies 4(gh-af)^2 - 4(g^2-ac)(h^2-ab) = 0[/math]

Simplify this and we will get

[math]\boxed{af^2+ch^2+bg^2-2fgh-abc = 0}[/math]

Hence this condition should be true if you want to factorize any such equation into two linear factor.
More neat and aesthetic form to express this condition is

[math] \begin{vmatrix} a & h & g \\ h & b & f \\ g & f & c \end{vmatrix} = 0 [/math].

This is also called discriminant of Conic section . (It decides whether the given equation represent an ellipse or parabola or circle or point or hyperbola or pair of straight line. Here it represents the pair of straight line).

Now come to this particular equation. (I am changing [math]a[/math] to [math]P[/math] and [math]b[/math] to [math]Q[/math] to avoid confusion).

So given equation is

[math]3x^2 + 2Qxy + 2y^2 + 2Px - 4y + 1[/math].

If you compare this to standard form you will find that ,

[math]a= 3 , h = Q , g = P , b = 2 , f = -2[/math] and [math]c = 1[/math]

So discriminant of this equation is

[math] \begin{vmatrix} 3 & P & Q \\ Q & 2 & -2 \\ P & -2 & 1 \end{vmatrix} = 0[/math]

Solving and rearranging this we have ,

[math]Q^2+4PQ + 2P^2 + 6 = 0 [/math]

Clearly [math]q[/math] is a root of quadratic equation given by

[math]x^2+4Px+2P^2+6[/math]