Q. Hello just wondering how I would go about getting a job in a big store like tesco, adsa, toys r us etc...
Do i just go in the store and ask at the kiosk for an application form? fill it in and send it and go to an interview if they call me?
Do i just go in the store and ask at the kiosk for an application form? fill it in and send it and go to an interview if they call me?
A. Absolutely....unless you want a Management Job, then you should jump through the proper hoops....service jobs are the easiest jobs to get today, as those are the only ones left out there.
Toys R Us Orientation.?
Q. Not to sound rude, but please dont answer with a question or answer about general Orientations. I'm asking about Toys R Us ones, not anywhere else.
So, I was scheduled for a group interview last Thursday, but had a midterm and couldn't make it. For some reason, I got called in for Orientation for this Thursday. I'm guessing they liked my application or didn't really notice I wasn't there, but i've read how the interviews are on Glassdoor.
Anyhow, now I'm just confused what to wear. I'm not sure if they have out the shirts yet, but I heard they usually do it at orientation. Should I wear black slacks and a black polo? Should it be tucked in? They also asked for me to bring two forms of identification, I only have one, will my passport suffice?
From what i've read of other answers similar to my question, itll be almost like a work day after we watch a training video. If I get paid, how would it happen?
Please and thanks!
Btw, if you were an employee for toysrus before, that'd be really great help.
So, I was scheduled for a group interview last Thursday, but had a midterm and couldn't make it. For some reason, I got called in for Orientation for this Thursday. I'm guessing they liked my application or didn't really notice I wasn't there, but i've read how the interviews are on Glassdoor.
Anyhow, now I'm just confused what to wear. I'm not sure if they have out the shirts yet, but I heard they usually do it at orientation. Should I wear black slacks and a black polo? Should it be tucked in? They also asked for me to bring two forms of identification, I only have one, will my passport suffice?
From what i've read of other answers similar to my question, itll be almost like a work day after we watch a training video. If I get paid, how would it happen?
Please and thanks!
Btw, if you were an employee for toysrus before, that'd be really great help.
A. You need two forms of identification that would be like your passport and a drivers id or a library card or a student id or a birth certificate or a voter id card or any other thing with your name on it
Black pants and a white shirt tucked in. you will get paid by a check or it will be by direct deposit in a checing account if you have one.
Black pants and a white shirt tucked in. you will get paid by a check or it will be by direct deposit in a checing account if you have one.
What makes complex numbers important?
Q. a + bi... My professor said something about CAT scans and e^x being related to sin(x) and cos(x) functions. He only touched upon this subject and then drifted back to integration techniques. Can someone briefly explain what imaginary and complex numbers are (Euler's formula derivation)? I'm only in calc II so please explain in layman's terms.
We haven't learned series (Taylor) yet. I should probably wait until I know how to do that before diving into this kind of math.
We haven't learned series (Taylor) yet. I should probably wait until I know how to do that before diving into this kind of math.
A. Beware: some parts here are definitely not for a layman but I guess you will understand them. In case not, you are welcome to ask for details. Here we go...
Complex numbers have one useful property that reals don't: every polynomial with complex coefficients is decomposable all the way to the root factors (x - root). This seemingly toy theorem has very significant corollaries and I would say that almost all applications of complex numbers is linked to it.
For example, it's known that complex numbers make a great aid in solving some important kinds of differential equations. The simplest case where they are needed, is:
a*y'' + b*y' + c*y = 0.
Theory then says we should look for special solutions in the form y = exp(λ*x). These are nice because they have simple derivatives:
y = exp(λx)
y' = λ exp(λx)
y'' = λ^2 exp(λx)
However, of course, only for some λ this is a solution. So, let's plug it in:
a*λ^2*exp(λx) + b*λ*exp(λx) + c*exp(λx) = 0
We can factor out the common term (which is surely nonzero):
a*λ^2 + b*λ + c = 0
So the derivatives of y became powers of λ.
Now this is a quadratic equation: you find some λ's using usual techniques and you immediately have some solutions of the original differential equation. So we know there are
* 2 solutions if b^2 - 4ac > 0
* 1 double root if b^2 - 4ac = 0 (this would be a bit more complicated, but let's not get in the details)
* no real solution if b^2 - 4ac < 0.
However, in the last case, we can use the quadratic formula unmodified if we allow complex numbers (and thus square roots of negative reals). So this gives also two roots λ, which are now imaginary (they have nonzero b in the form a + bi -- not to be confused with the coefficients a,b,c above!), and complex conjugate to each other (that is, they are a + bi and a - bi).
Here the Euler's formula comes into play, which tells that
exp(i*x) = cos(x) + i*sin(x).
[[N.B. exp(x) is the same as e^x.]]
The exponential function in complex numbers is obtained by a rather complicated process called analytical continuation. See below for details. However, it preserves all its nice properties except from that it was an injection. So, for example, the following multiplication rule holds:
exp(a + bi) = exp(a) exp(bi) = exp(a) * (cos(b) + i*sin(b))
So the two solutions of the DE can be written as
y1(x) = exp(λ1*x) = exp((a+bi)*x) = exp(ax) * (cos(bx) + i*sin(bx))
and
y2(x) = exp(λ2*x) = exp((a-bi)*x) = exp(ax) * (cos(-bx) + i*sin(-bx)) = exp(ax) * (cos(bx) - i*sin(bx))
The theory then says that if we have some set of solutions, any of their linear combinations is again a solution. So we can for example do
y3(x) = 1/2 * (y1(x) + y2(x)) = exp(ax) * cos(bx)
y4(x) = 1/(2i) * (y1(x) - y2(x)) = exp(ax) * sin(bx),
which returns us back to real functions of one real variable. Using this as a new so-called fundamental system, we can write a general solution of the DE as
exp(ax) * (A*cos(bx) + B*sin(bx)) with parameters A, B
or
C * exp(ax) * cos(bx + �0) with parameters C, �.
Therefore, we have solved a real equation with real unknown functions using complex numbers as an aid, and effectively exploiting their property that the polynomial ax^2 + bx + c can be solved regardless of the sign of its discriminant.
I hope this example clarifies your question a bit...
On Euler formula derivation:
The way I know for this uses so-called Taylor expansions of functions. To follow me, you have to believe that it holds that
exp(x) = sum [n = 0 to +�] x^n / n!
sin(x) = sum [n = 0 to +�] (-1)^n / (2*n+1)! * x^(2*n+1)
cos(x) = sum [n = 0 to +�] (-1)^n / (2*n)! * x^(2*n)
for each x in R.
We can take the first formula and simply say that we extend its validity to complex x's as well. Let's see then what that does if we plug i*x in the place of x:
exp(i*x) = sum [n = 0 to +�] (i*x)^n / n! = sum [n = 0 to +�] i^n * x^n / n!
So a term of i^n appeared now. That is:
* i^0 = 1
* i^1 = i
* i^2 = -1
* i^3 = -i
* i^4 = 1 again, so we can say:
** i^n = 1 for n = 0, 4, 8, 12, ...
** i^n = i for n = 1, 5, 9, ...
** i^n = -1 for n = 2, 6, 10, ...
** i^n = -i for n = 3, 7, 11, ...
We can handle the minuses easily but we will at least separate even n's from odd ones to split the real part (where i^n is a real number) and the imaginary part (where it is +i or -i). Therefore
* if n = 2*k, i^n * x^n / n! = (-1)^k * x^(2k) / (2k)!
* if n = 2*k+1, i^n * x^n / n! = (-1)^k * i * x^(2k+1) / (2k+1)!
If we let k go from 0 to +�, these two cases will reach each possible n. Finally,
exp(i*x) = sum [n = 0 to +�] (i*x)^n / n! = sum [n = 0 to +�] i^n * x^n / n! = sum [k = 0 to +�] (-1)^k * x^(2k) / (2k)! + i * sum [k = 0 to +�] (-1)^k * x^(2k+1) / (2k+1)!
which is nothing else than cos(x) + i*sin(x).
Complex numbers have one useful property that reals don't: every polynomial with complex coefficients is decomposable all the way to the root factors (x - root). This seemingly toy theorem has very significant corollaries and I would say that almost all applications of complex numbers is linked to it.
For example, it's known that complex numbers make a great aid in solving some important kinds of differential equations. The simplest case where they are needed, is:
a*y'' + b*y' + c*y = 0.
Theory then says we should look for special solutions in the form y = exp(λ*x). These are nice because they have simple derivatives:
y = exp(λx)
y' = λ exp(λx)
y'' = λ^2 exp(λx)
However, of course, only for some λ this is a solution. So, let's plug it in:
a*λ^2*exp(λx) + b*λ*exp(λx) + c*exp(λx) = 0
We can factor out the common term (which is surely nonzero):
a*λ^2 + b*λ + c = 0
So the derivatives of y became powers of λ.
Now this is a quadratic equation: you find some λ's using usual techniques and you immediately have some solutions of the original differential equation. So we know there are
* 2 solutions if b^2 - 4ac > 0
* 1 double root if b^2 - 4ac = 0 (this would be a bit more complicated, but let's not get in the details)
* no real solution if b^2 - 4ac < 0.
However, in the last case, we can use the quadratic formula unmodified if we allow complex numbers (and thus square roots of negative reals). So this gives also two roots λ, which are now imaginary (they have nonzero b in the form a + bi -- not to be confused with the coefficients a,b,c above!), and complex conjugate to each other (that is, they are a + bi and a - bi).
Here the Euler's formula comes into play, which tells that
exp(i*x) = cos(x) + i*sin(x).
[[N.B. exp(x) is the same as e^x.]]
The exponential function in complex numbers is obtained by a rather complicated process called analytical continuation. See below for details. However, it preserves all its nice properties except from that it was an injection. So, for example, the following multiplication rule holds:
exp(a + bi) = exp(a) exp(bi) = exp(a) * (cos(b) + i*sin(b))
So the two solutions of the DE can be written as
y1(x) = exp(λ1*x) = exp((a+bi)*x) = exp(ax) * (cos(bx) + i*sin(bx))
and
y2(x) = exp(λ2*x) = exp((a-bi)*x) = exp(ax) * (cos(-bx) + i*sin(-bx)) = exp(ax) * (cos(bx) - i*sin(bx))
The theory then says that if we have some set of solutions, any of their linear combinations is again a solution. So we can for example do
y3(x) = 1/2 * (y1(x) + y2(x)) = exp(ax) * cos(bx)
y4(x) = 1/(2i) * (y1(x) - y2(x)) = exp(ax) * sin(bx),
which returns us back to real functions of one real variable. Using this as a new so-called fundamental system, we can write a general solution of the DE as
exp(ax) * (A*cos(bx) + B*sin(bx)) with parameters A, B
or
C * exp(ax) * cos(bx + �0) with parameters C, �.
Therefore, we have solved a real equation with real unknown functions using complex numbers as an aid, and effectively exploiting their property that the polynomial ax^2 + bx + c can be solved regardless of the sign of its discriminant.
I hope this example clarifies your question a bit...
On Euler formula derivation:
The way I know for this uses so-called Taylor expansions of functions. To follow me, you have to believe that it holds that
exp(x) = sum [n = 0 to +�] x^n / n!
sin(x) = sum [n = 0 to +�] (-1)^n / (2*n+1)! * x^(2*n+1)
cos(x) = sum [n = 0 to +�] (-1)^n / (2*n)! * x^(2*n)
for each x in R.
We can take the first formula and simply say that we extend its validity to complex x's as well. Let's see then what that does if we plug i*x in the place of x:
exp(i*x) = sum [n = 0 to +�] (i*x)^n / n! = sum [n = 0 to +�] i^n * x^n / n!
So a term of i^n appeared now. That is:
* i^0 = 1
* i^1 = i
* i^2 = -1
* i^3 = -i
* i^4 = 1 again, so we can say:
** i^n = 1 for n = 0, 4, 8, 12, ...
** i^n = i for n = 1, 5, 9, ...
** i^n = -1 for n = 2, 6, 10, ...
** i^n = -i for n = 3, 7, 11, ...
We can handle the minuses easily but we will at least separate even n's from odd ones to split the real part (where i^n is a real number) and the imaginary part (where it is +i or -i). Therefore
* if n = 2*k, i^n * x^n / n! = (-1)^k * x^(2k) / (2k)!
* if n = 2*k+1, i^n * x^n / n! = (-1)^k * i * x^(2k+1) / (2k+1)!
If we let k go from 0 to +�, these two cases will reach each possible n. Finally,
exp(i*x) = sum [n = 0 to +�] (i*x)^n / n! = sum [n = 0 to +�] i^n * x^n / n! = sum [k = 0 to +�] (-1)^k * x^(2k) / (2k)! + i * sum [k = 0 to +�] (-1)^k * x^(2k+1) / (2k+1)!
which is nothing else than cos(x) + i*sin(x).
Everytime I try to apply online..?
Q. they always ask for my social security number!
How do people get these jobs and not put in their social security number?
:(
I tried to get the paper form of the application but they ALL said to apply online. And I went to about 6-7 different places.
Is it safe to put your social security number on these websites? Like.. CVS, or Toys R us?
How do people get these jobs and not put in their social security number?
:(
I tried to get the paper form of the application but they ALL said to apply online. And I went to about 6-7 different places.
Is it safe to put your social security number on these websites? Like.. CVS, or Toys R us?
A. If you can avoid it, leave out your SSN, but indicate on the application that if you are called for interview, you will be happy to provide it for a background & credit check - because that is what the SSN is supposed to be for.
Putting your SSN on an initial application is very dangerous because you never know who will be reviewing the application, or where it will be thrown or filed if you are not in the running.
I have seen jpb applications in the trash can of large companies - and if someone were to pick it up..... well, you know what will likely happen.
Putting your SSN on an initial application is very dangerous because you never know who will be reviewing the application, or where it will be thrown or filed if you are not in the running.
I have seen jpb applications in the trash can of large companies - and if someone were to pick it up..... well, you know what will likely happen.
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