Wotzdorfg
2021-03-07
Answered

Having difficulties with signed numbers when it comes to multiplication, division, addition and subtraction. Is there a simpler way to remember how to use the signs in math problems?

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Ayesha Gomez

Answered 2021-03-08
Author has **104** answers

Let two negative numbers be -a and -b. For example : -2 and -5.

Let us consider the rules for individual arithmatic operations:

Addition of 2 negatuve numbers:

e.g,

Substraction of 2 negative numbers:

e.g,

Multiplication of 2 negative numbers:

e.g.,

Division of 2 negative numbers:

e.g,

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Find an equation of the tangent plane to the given surface at the specified point. $z=3{y}^{2}-2{x}^{2}+x,\text{}(2,-1,-3)$

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A set of ordered pairs is called a _______.

asked 2020-11-07

Show that the least upper bound of a set of negative numbers cannot be positive.

asked 2021-04-06

Find the point on the plane $x+2y+3z=13$ that is closest to the point (1,1,1). How would you minimize the function?

asked 2022-07-02

In measure theory I encountered Egorov's theorem which states that if $(X,\mathcal{S},\mu )$ is a measure space such that $\mu (X)<\mathrm{\infty}$ i.e. $\mu $ is a finite measure.If $({f}_{n})$ be a sequence of measurable functions on $X$ converging pointwise to $f:X\to \mathbb{R}$,then ${f}_{n}$ is almost uniformly convergent to $f$.

Now in the book the definition of almost uniform convergence is the following:

${f}_{n}\to f$ almost uniformly if for each $\u03f5>0$,there exists ${E}_{\u03f5}\subset X$ such that $\mu ({E}_{\u03f5})<\u03f5$ and ${f}_{n}\to f$ uniformly on $X-{E}_{\u03f5}$.

Now in some books I have seen that almost everywhere means outside a measure 0 set.

So the definition of almost uniformly convergent should have been ${f}_{n}\to f$ almost uniformly if $\mathrm{\exists}E\subset X$ such that $\mu (E)=0$ and ${f}_{n}\to f$ uniformly on $X-E$.

But unfortunately the definition is not so.In fact the latter condition is stronger.I want to know why the former is taken as a definition and what the problem with the latter one is.I want to understand where I am making mistake in understanding the word "almost".

Now in the book the definition of almost uniform convergence is the following:

${f}_{n}\to f$ almost uniformly if for each $\u03f5>0$,there exists ${E}_{\u03f5}\subset X$ such that $\mu ({E}_{\u03f5})<\u03f5$ and ${f}_{n}\to f$ uniformly on $X-{E}_{\u03f5}$.

Now in some books I have seen that almost everywhere means outside a measure 0 set.

So the definition of almost uniformly convergent should have been ${f}_{n}\to f$ almost uniformly if $\mathrm{\exists}E\subset X$ such that $\mu (E)=0$ and ${f}_{n}\to f$ uniformly on $X-E$.

But unfortunately the definition is not so.In fact the latter condition is stronger.I want to know why the former is taken as a definition and what the problem with the latter one is.I want to understand where I am making mistake in understanding the word "almost".

asked 2022-05-21

Simple question: Let $(E,\mathcal{E})$ be a measurable space and $\mathrm{\Delta}\notin E$. Now let ${E}_{\mathrm{\Delta}}:=E\cup \{\mathrm{\Delta}\}$ and ${\mathcal{E}}_{\mathrm{\Delta}}$ denote the smallest σ-algebra on ${E}_{\mathrm{\Delta}}$ containing $\mathcal{E}$.

Question: Can we given an explicit formula for ${\mathcal{E}}_{\mathrm{\Delta}}$?

We've clearly got $\{\mathrm{\Delta}\}\in {\mathcal{E}}_{\mathrm{\Delta}}$, since $E,{E}_{\mathrm{\Delta}}\in {\mathcal{E}}_{\mathrm{\Delta}}$.

BTW: Is there a better, established notation for the $\sigma $-algebra generated by $\mathcal{E}$ on ${E}_{\mathrm{\Delta}}$? Writing $\sigma (\mathcal{E})$ wouldn't make clear whether we mean the $\sigma $-algebra generated on $E$ or a larger space like ${E}_{\mathrm{\Delta}}$.

Question: Can we given an explicit formula for ${\mathcal{E}}_{\mathrm{\Delta}}$?

We've clearly got $\{\mathrm{\Delta}\}\in {\mathcal{E}}_{\mathrm{\Delta}}$, since $E,{E}_{\mathrm{\Delta}}\in {\mathcal{E}}_{\mathrm{\Delta}}$.

BTW: Is there a better, established notation for the $\sigma $-algebra generated by $\mathcal{E}$ on ${E}_{\mathrm{\Delta}}$? Writing $\sigma (\mathcal{E})$ wouldn't make clear whether we mean the $\sigma $-algebra generated on $E$ or a larger space like ${E}_{\mathrm{\Delta}}$.

asked 2022-07-18

Solving inverse trig question without forming cases

Question:

Find $x$ if $\mathrm{arctan}(x+3)-\mathrm{arctan}(x-3)=\mathrm{arctan}(3/4)$

My attempt:

I know the formula:

$\mathrm{arctan}x-\mathrm{arctan}y=\mathrm{arctan}(\frac{x-y}{1+xy})$

for both $x,y>0$

If both $x,y$ are NOT greater than $0$ then we'll need to form a second case in the above formula.

I solved assuming both $x+3$ and $x-3$ are greater than zero and got a quadratic in the end solving which I got $x=\pm 4$

I dutifully rejected $x=-4$ as it invalidated my assumption. However on putting in the above equation I found that it DOES satisfy the equation.

I do not wish to form cases for positive/negative. I wish to know if there is a simpler and direct way to solve such a question. Thanks!

Question:

Find $x$ if $\mathrm{arctan}(x+3)-\mathrm{arctan}(x-3)=\mathrm{arctan}(3/4)$

My attempt:

I know the formula:

$\mathrm{arctan}x-\mathrm{arctan}y=\mathrm{arctan}(\frac{x-y}{1+xy})$

for both $x,y>0$

If both $x,y$ are NOT greater than $0$ then we'll need to form a second case in the above formula.

I solved assuming both $x+3$ and $x-3$ are greater than zero and got a quadratic in the end solving which I got $x=\pm 4$

I dutifully rejected $x=-4$ as it invalidated my assumption. However on putting in the above equation I found that it DOES satisfy the equation.

I do not wish to form cases for positive/negative. I wish to know if there is a simpler and direct way to solve such a question. Thanks!