php – How do I make product cards the same height?

I am a beginner in all of this and I try to make these product cards the same. I've browsed through several websites and posts that tried to edit but to no avail. I think the problem is the different number of lines in the title of the products – the titles that have the same number of lines are the same height. However, I have to have all product cards the same. What can I do?

My code:

                      

Styling:

#content {
   padding-left: 25px; 
}

.single {
    width: 290px;
}

@media(max-width:768px) {
    .single{
        width: 60%;
        margin: 0 auto;
    }
}

#content .product {
    background: #ffffff;
    border: solid 1px #e6e6e6;
    margin-bottom: 30px;
    box-sizing: border-box;
}

#content .product .text {
    padding: 10px 10px 0px;
}

#content .product .text p.price {
    font-size: 18px;
    text-align: center;
    font-weight: 400;
}

#content .product .text .button {
    text-align: center;
    clear: both;
}

#content .product .text .button .btn {
    margin-bottom: 10px;
}

#content .product .text h3 {
    text-align: center;
    font-size: 20px;
}

#content .product .text h3 a {
    color: rgb(85, 85, 85);
}

Everything on the website is included in a

What i get
What I want

If anyone could let me know how to add CSS styling properly (unless that's the right way) I would very much appreciate it.

Thanks a lot!

real analysis – can we get a Sobolev card with a.e. fixed singular values ​​due to smooth cards with fixed singular values?

To let $ 0 < sigma_1 < sigma_2 $, and let $ D subseteq mathbb {R} ^ 2 $ be the closed unit disk.

To let $ f in W ^ {1,2} (D, mathbb {R} ^ 2) $and assume that the singular values ​​of $ df $ are a.e. like $ sigma_1, sigma_2 $,

Is there $ f_n in C ^ { infty} (D ^ o, mathbb {R} ^ 2) $ so that $ f_n to f $ in the $ W ^ {1,2} $ and the singular values ​​of $ df_n $ are the same everywhere $ sigma_i $?

Let us assume that $ f $ is continuous, we can also get $ f_n $ which converge evenly $ f $?

The standard mollification process doesn't work here – it's an averaging process, and averaging matrices with given singular values ​​reduces the norm. (All of these matrices lie on the Euclidean sphere of the radius $ sqrt { sigma_1 ^ 2 + sigma_2 ^ 2} $which is strictly convex).

Note that there are no affine smooth cards with fixed singular values.


Here is a concrete example:

Suppose that $ frac { sigma_2} { sigma_1} = n $ is a natural number.

The map $ f: re ^ {i theta} to sigma_1re ^ {i ( sigma_2 / sigma_1) theta} $ (which is generally only smoothly defined on the plate after removing a beam) has constant singular values $ sigma_i $,

We assumed that $ frac { sigma_2} { sigma_1} = n $.
$$
f (z) = f (re ^ {i theta}) = re ^ {in theta} = frac {z ^ n} {| z | ^ {n-1}},
$$

is smooth on the entire hard drive without the origin and is in $ W ^ {1,2} (D, mathbb {R} ^ 2) $,

Can we get closer? $ f $ in the $ W ^ {1,2} $ with smooth cards with the fixed singular values $ sigma_i $?

fa.functional analysis – Completely positive, uniform cards that affect uniform operators

Consider a completely positive, consistent, and primitive card $ E $ on the operators in a finite-dimensional Hilbert space and assume that there is a uniform and a Hermitian operator $ U $ (therefore $ U ^ 2 = 1 $), which is itself assigned to a uniform (and Hermitian) operator of $ E $,
If it is assigned to a multiple of itself, then $ U = pm 1 $ Let us assume that the primitiveness is assigned to a unit that is not proportional to the identity.
My basic question is whether this is possible (i.e. an example of such a card) $ E $ would be happy to prove it $ U $ should be $ pm 1 $).

Note that we have this through equality in cadison black $ E (UX) = E (U) E (X) $ and $ E (XU) = E (X) E (U) $ for all $ X $, Since $ U ^ 2 = 1 $we have further $ U = P-Q $With $ P $ and $ Q $ orthogonal projectors so that $ 1 = P + Q $, It is not very difficult to deduce that $ E (P) ^ 2 = E (P) $ using equality in the Kadison-Black inequality and similar for $ Q $, It also follows from this $ E (PX) = E (P) E (X) $ and similar for $ Q $ (and the reverse order). If you could show $ E (P) leq P $, then none $ PYP $ would be mapped $ PE (PYP) P $, Since $ E $ is primitive and therefore irreducible, would imply that $ P in {1,0 } $ and therefore $ U = pm 1 $,

It is easy to create examples of non-primitive, unitary CP maps $ E (P) leq P $ hurt, but somehow I haven't made it to primitive, although I wouldn't be very surprised if there is a simple counterexample.

Any insight would be greatly appreciated!
I asked this question earlier during the batch exchange, but maybe I'm lucky here. It may not be surprising that the problem appeared in an argument from quantum information theory, but it has now become an independent problem 🙂

How can I prevent cards from being repeated in the same Anki session?

When I click on "Again" while studying a card, I want it not to be presented again in the same session, even if the session lasts about 10 minutes, which is longer than the shortest repetition time, about 1 minute. (I use Anki 2.1.)

For example, when I enter 100 new cards, I want to be able to open a session and go through all the cards that mark each card as "again," even though I remember most of them well. Close the session and repeat the process one or more times later in the day or the next day before continuing with normal use of Anki. This is what I like to do right from the start, as a kind of informal reminder session or test in which each card is presented exactly once and each card is left at the lowest level for the next session. It is annoying at this point in time when some cards are presented twice or more, as it can take a long time before the last cards are presented to me.

I suppose a workaround would be to set the first learning step longer than the session length, but I was wondering if there is a method that does not require changing the learning step length.

I should add that minutes don't mean a lot to me when it comes to Anki because there is no device that I run on, that I keep on, or that I want Anki to catch my eye, even if it is on. I plan to use Anki a few times a day when I have the opportunity to, and make cards in stacks of about 100 and check those cards a few times as described in the second paragraph before doing Anki leave the distance of repetition.

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fa.functional analysis – When do completely positive cards have a closed picture?

To let $ mathcal {A}, mathcal {B} $ Be C * algebras. A map $ phi colon mathcal {A} rightarrow mathcal {B} $ is completely positive (cp) if it is linear, * get it and all of its & # 39; coordinate extensions to matrices $ phi ^ {(n)} Colon M_n ( mathcal {A}) rightarrow M_n ( mathcal {B}) $ Map positive matrices to positive matrices. For example, according to Stinesprings & # 39; representation, completely positive cards are only in shape $ V pi V ^ * $, Where $ pi $ is a * homomorphism.

My question is under what conditions these CP cards have a closed picture (see here for the proof that homomorphisms have a closed picture).

This is true, of course, if the card is isometric, but I can't prove that injective norm 1 CP cards, for example, are isometric. The usual evidence for homomorphisms is that the spectrum remains unchanged under a homomorphism. Does this also apply to CP maps? It seems wrong to me.

Thanks for any help you can give. I have to overlook something.