Which number is rational?

Mathematics

1 answer:

ziro4ka [17]3 years ago

5 0

Answer:

We need some numbers to work with to truly answer this. However here's a definition of a rational numbers

A rational number is any number that isn't infinite and non-repeating. So decimals, fractions, repeating numbers/decimals like 1.535353.... (53 is the "pattern") are all rational and fair game. Numbers like pi (π), "imperfect" square roots, etc. that don't repeat and are infinite or NOT RATIONAL.

<u>Hope this helps and have a nice day!</u>

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Six decreased by 9 times a number is -39

Andrew [12]

The translation of the mathematical statement given as six decreased by 9 times a number is -39 is 9x - 6 = -39

<h3>How to translate the mathematical statement?</h3>

The mathematical statement is given as:

six decreased by 9 times a number is -39

9 times a number means

9 * x (assume x is the number)

So, we have:

six decreased by 9 times a number is -39 ⇒ six decreased by 9x is -39

Decreased by means we subtract 6 from 9x.

So, we have:

six decreased by 9 times a number is -39 ⇒ 9x - 6 = -39

Hence, the translation of the mathematical statement given as six decreased by 9 times a number is -39 is 9x - 6 = -39

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1 year ago

What is the prime factorization of 40?<br> 1. 2x5<br> 2. 2x20<br> 3. 2x2x2x5<br> 4. 2x2x4x5

noname [10]

Answer: number 3 is the answer

3 0

3 years ago

Read 2 more answers

Use chain rule to find dw/dt w=ln(x^2 + y^2 + z^2)^(1/2) , x=sint, y=cost, z=tant

son4ous [18]

The chain rule states that

$\dfrac{\mathrm dw}{\mathrm dt}=\dfrac{\partial w}{\partial x}\dfrac{\mathrm dx}{\mathrm dt}+\dfrac{\partial w}{\partial y}\dfrac{\mathrm dy}{\mathrm dt}+\dfrac{\partial w}{\partial z}\dfrac{\mathrm dz}{\mathrm dt}$

Since $\ln(x^2+y^2+z^2)^{1/2}=\dfrac12\ln(x^2+y^2+z^2)$ , you have

$\dfrac{\partial w}{\partial x}=\dfrac x{x^2+y^2+z^2}$
$\dfrac{\partial w}{\partial x}=\dfrac y{x^2+y^2+z^2}$
$\dfrac{\partial w}{\partial z}=\dfrac z{x^2+y^2+z^2}$

and

$\dfrac{\mathrm dx}{\mathrm dt}=\cos t$
$\dfrac{\mathrm dy}{\mathrm dt}=-\sin t$
$\dfrac{\mathrm dz}{\mathrm dt}=\sec^2t$

Also, since $x^2+y^2+z^2=\sin^2t+\cos^2t+\tan^2t=1+\tan^2t=\sec^2t$ , the derivative is

$\dfrac{\mathrm dw}{\mathrm dt}=\dfrac{\sin t\cos t}{\sec^2t}-\dfrac{\sin t\cos t}{\sec^2t}+\dfrac{\tan t\sec^2t}{\sec^2t}$
$\dfrac{\mathrm dw}{\mathrm dt}=\tan t$