1st grade mathematics work sheets and my mother's mathematics teaching style. Mathematics will not be so terrible, since it seems that parents are interested in preparing their little ones for mathematics before school age.

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You can always keep track of things like your money and solve certain real-life problems by putting things in an equation and then resolving them.

Knowing the algebra is something that people should do to achieve it to find a good job and for success in college, as well as life. Learn mathematics such as algebra and arithmetic now because as time passes, it becomes more difficult to learn things.

Math action like drill and practice help a kid increasing great mental number aptitudes. In addition, if the training is done regularly through playing around, youngsters effectively get acquainted with numbers.

These days, we have various online numerical exercises implied for kids and the adults the same. Instructors, guardians and youngsters have various games and number drill exercises close by.

The key to teaching division to young students who have only been introduced recently to more complicated mathematics forms is to make the student feel involved with the process, while finding it pleasant at the same time.

Mathematics worksheets are effective, but there are other ways in which mathematics can effectively teach. This is practically the generic form of teaching that demonstrates being more effective among children and young people.

No matter how many times make the child complete a division worksheet, or produces division assignments, there is no promise that this child can deliver continuously with the ability to perform successful division problems.

Children these days are expected to do more at school at an earlier age than ever. Years ago, students were learning to read and write at first grade. Nowadays, children are learning to read and write in kindergarten.

With the increase in the demands of the school curriculum, it is not surprising that the subject of "handwriting" or "penmanship" has fallen by the way.

@DesmondSwayne Sir Desmond is right. SAGE makes mathematical predictions based on flawed models.
I chose to stick my neck out almost 4mo ago here:
https://t.co/b0rT5Lq9HI
I experienced much anxiety doing this, aware that, if I was badly wrong here, I would be destroyed as a contributor.

That said, I thought the immunology so clear that I made three, specific & testable predictions which should apply if I was right.
Please examine them, on the right, under the pie chart:

In particular, I predicted that the epidemic would not reignite in London. Obviously, we have many “cases” but I urge people to disregard these & instead to look at excess deaths, on the principle that a severe (occasionally lethal) respiratory virus would show itself this way.

Here is a chart showing the trend of deaths in London alone over recent years. A prominent peak of deaths is evident as the virus swept through in spring. Also visible the August heatwave. But there are no excess deaths now.
I submit this is a striking example of herd immunity.

I’m still waiting for SAGE or indeed anyone to provide an alternative explanation for the sharp nature of the spring excess deaths peak which doesn’t involve herd immunity or otherwise removing from the pool those vulnerable to lethal outcomes (same result). Given this sharp...

When I was starting out as a grad student, I remember being frustrated with analysis problems where it didn't use DIRECTLY the things we spent so much time proving in class.

As an example, let's look at a standard real analysis problem:

(1/23)

The problem looks like it can be done using the dominated convergence theorem.

This is an important theorem and is used in a lot of places outside a grad real analysis class. A student would go through a lot of buildup leading to this theorem.

(2/23)

Now, let's see whether the conditions for DCT are satisfied.

First, for fixed 00 as n->infinity.

So we have a.e. convergence of the integrand in [0,1].

To the student, this builds a stronger case to use DCT to solve the problem.

(3/23)

Finding an integrable dominating function is where the problem starts.

It would seem NATURAL (at least for me) to bound it with:

p.s. it feels "natural" to me because it is reminiscent of squeeze theorem problems, math students encounter in calculus.

(4/23)

However, 1/x^n is not integrable in (0,1) and hence cannot be (further) dominated by an integrable function.

So, what I would do is since I know it converges a.e. to 1, maybe it converges uniformly to 1?

This leads me into a BAD detour.

(5/23)

⬇️⬇️⬇️ The "Math" of Fear

As it is Thursday, I have a post for you today in anticipation of Gov. DeWine's new Maps of Fear which he has tried to phase in. I have previously discussed both the new red and blue maps and their issues -

- but today I hope to illustrate even more clearly what level of manipulation is going on, particularly with the new blue 'ICU Utilization' map.

Attached I have two images.

The first is a hyper-simplistic (and unrealistic) but illustrative example of the calculations that the new blue map goes through to arrive at its numbers.

Instead of telling us the percent of beds occupied by COVID+ (not necessarily ill *with* COVID, but testing positive) - the governor and ODH only calculates the percentage of COVID+ out of just the *used* beds.

In the first image, I have an example of a 100 bed ICU with just two patients - one COVID positive (red) and one COVID negative (green). The other 98 beds are empty and able to accept new patients.

How to mentally compute the cube root of the cube of a two digit number.

Thread (1/10)

In base 10, the last digits of the cubes of digits are distinct.

If d = 0, 1, 4, 5, 6, or 9, then d^3 ends in d.

If d = 2, 3, 7, or 8, then d^3 ends in 10-d.

(2/10)

So if you're given the cube of an integer, you can figure out the last digit of the cube root.

For example, if 50653 is the cube of an integer, it's the cube of a number ending in 7.

(3/10)

Let n = 10a + b be a two-digit number, with a and b single digit numbers.

Then 10a <= n <= 10(a+1), and so n^3 is between (10a)^3 and (10(a+1))^3, i.e. between 1000 a^3 and 1000 (a+1)^3.

(4/10)

So suppose x is the cube of a two-digit number n = 10a + b.

Chop off the last three digits of x.

Then a, the first digit of n, is the largest number whose cube is no greater than what's left of x.

(5/10)