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2 years ago

Read the following two statements. Then use the Law of Syllogism to draw a conclusion.

Mathematics

2 answers:

Tema [17]2 years ago

6 0

Answer: A. If Wally goes to the mall, then Alice will go to the mall.

Step-by-step explanation:

The given conditional statements :

If Frita goes to the mall, then Alice will go to the mall.

If Wally goes to the mall, then Frita will go to the mall.

We can see that the hypothesis in the first statement and the conclusion in the second statement is same.

Therefore, by the law of syllogism, the conclusion statement will be:

" If Wally goes to the mall, then Alice will go to the mall."

trapecia [35]2 years ago

5 0

A. If Wally goes to the mall, then Alice will go to the mall.

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WINSTONCH [101]

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2 years ago

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3 years ago

Find the area of the regular 2424-gon with radius 1010 mm.

Mkey [24]

Answer:

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Step-by-step explanation:

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2 years ago

Give me an example of an irrational number that is greater than 10

Pavel [41]

To find:

An irrational number that is greater than 10.

Solution:

Irritation number: It cannot be expression in the form of $\dfrac{p}{q}$ , where, $q\neq 0, p,q$ are integers.

For example: $\sqrt{2}\sqrt{3},\pi, 1.263689...,etc.$ .

We know that square of 10 is 100. So, square root of any prime number is an example of an irrational number that is greater than 10.

First prime number after 100 is 101.

Required irrational number $=\sqrt{101}$

Therefore, $\sqrt{101}$ is an irrational number that is greater than 10.

5 0

3 years ago

For <img src="https://tex.z-dn.net/?f=e%5E%7B-x%5E2%2F2%7D" id="TexFormula1" title="e^{-x^2/2}" alt="e^{-x^2/2}" align="absmiddl

nevsk [136]

I'm assuming you're talking about the indefinite integral

$\displaystyle\int e^{-x^2/2}\,\mathrm dx$

and that your question is whether the substitution $u=\dfrac x{\sqrt2}$ would work. Well, let's check it out:

$u=\dfrac x{\sqrt2}\implies\mathrm du=\dfrac{\mathrm dx}{\sqrt2}$
$\implies\displaystyle\int e^{-x^2/2}\,\mathrm dx=\sqrt2\int e^{-(\sqrt2\,u)^2/2}\,\mathrm du$
$=\displaystyle\sqrt2\int e^{-u^2}\,\mathrm du$

which essentially brings us to back to where we started. (The substitution only served to remove the scale factor in the exponent.)

What if we tried $u=\sqrt t$ next? Then $\mathrm du=\dfrac{\mathrm dt}{2\sqrt t}$ , giving

$=\displaystyle\frac1{\sqrt2}\int \frac{e^{-(\sqrt t)^2}}{\sqrt t}\,\mathrm dt=\frac1{\sqrt2}\int\frac{e^{-t}}{\sqrt t}\,\mathrm dt$

Next you may be tempted to try to integrate this by parts, but that will get you nowhere.

So how to deal with this integral? The answer lies in what's called the "error function" defined as

$\mathrm{erf}(x)=\displaystyle\frac2{\sqrt\pi}\int_0^xe^{-t^2}\,\mathrm dt$

By the fundamental theorem of calculus, taking the derivative of both sides yields

$\dfrac{\mathrm d}{\mathrm dx}\mathrm{erf}(x)=\dfrac2{\sqrt\pi}e^{-x^2}$

and so the antiderivative would be

$\displaystyle\int e^{-x^2/2}\,\mathrm dx=\sqrt{\frac\pi2}\mathrm{erf}\left(\frac x{\sqrt2}\right)$

The takeaway here is that a new function (i.e. not some combination of simpler functions like regular exponential, logarithmic, periodic, or polynomial functions) is needed to capture the antiderivative.

3 0

2 years ago