Hi Badinage,

I have not seen you for ages

I am glad that you are here now :)))

Thank you Melody.

If I correctly understand what you have in mind, then try replacing each decimal with its equivalent common fraction, e.g., instead of 3.86 you can write \({386} \over {100}\)

Does that help?

🐟 🐬

Very impressive Heureka :)

"how do you simplify 32 with an exponent of 3 over 5"

\(32^{3/5}\rightarrow(32^{1/5})^3\rightarrow2^3\rightarrow8\)

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You cannot obtain an explicit algebraic solution in the form x = something here, where the 'something' is independent of x.

A possible numerical approach is to use the Newton-Raphson method, where you have to subtract one side from the other to create a function that is equal to zero:

The positive solution here is x = 2.314.  The negative solution could be obtained by using an initial guess of, say -3, though the symmetry here makes it obvious that the negative solution is x = -2.314.

You would need to use a similar numerical method for your other suggestion.

Here one obvious solution is x = pi.  Use Newton-Raphson to find the others.  For example:

So another solution is x = 0.828 (and because of the sine curve, the final solution is at x = 2pi - 0.828).

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How using L'Hospital's rule?

\(\displaystyle\lim \limits_{x\to 0} \left(\frac{(1+x)^{ \frac{1}{x} }}{e} \right)^{ \frac{1}{x} }\)

Formula: \(\begin{array}{|rcll|} \hline \ln \left(\displaystyle\lim \limits_{x\to 0} f(x)\right) &=& \displaystyle\lim \limits_{x\to 0} \ln f(x) \\ \hline \end{array} \)

1. We  take the logarithm:

\(\begin{array}{|rcll|} \hline \ln \left(\displaystyle\lim \limits_{x\to 0} \left(\frac{(1+x)^{ \frac{1}{x} }}{e} \right)^{ \frac{1}{x} }\right) &=& \displaystyle\lim \limits_{x\to 0} \ln \left\{ \left(\frac{(1+x)^{ \frac{1}{x} }}{e} \right)^{ \frac{1}{x} } \right\}\\ &=& \displaystyle\lim \limits_{x\to 0} \frac{1}{x}\cdot \ln \left(\frac{(1+x)^{ \frac{1}{x} }}{e} \right)\\ &=& \displaystyle\lim \limits_{x\to 0} \frac{1}{x}\cdot \left[ \ln \left( (1+x)^{ \frac{1}{x} } \right) -\ln(e) \right] \quad &|\quad \ln(e) = 1\\ &=& \displaystyle\lim \limits_{x\to 0} \frac{1}{x}\cdot \left[ \ln \left( (1+x)^{ \frac{1}{x} } \right) -1 \right] \\ &=& \displaystyle\lim \limits_{x\to 0} \frac{1}{x}\cdot \left[ \frac{1}{x}\cdot\ln(1+x) -1 \right] \\ &=& \displaystyle\lim \limits_{x\to 0} \frac{1}{x}\cdot \left[ \frac{1}{x}\cdot\ln(1+x) -\frac{x}{x} \right] \\ &=& \displaystyle\lim \limits_{x\to 0} \frac{1}{x}\cdot \left[ \frac{\ln(1+x)-x}{x} \right] \\ &=& \displaystyle\lim \limits_{x\to 0} \frac{\ln(1+x)-x}{x^2} \quad &|\quad \text{L'Hospital's rule}\\ &=& \displaystyle\lim \limits_{x\to 0} \frac{(\ln(1+x)-x)'}{(x^2)'} \\ &=& \displaystyle\lim \limits_{x\to 0} \frac{ \frac{1}{1+x}-1 }{2x} \\ &=& \displaystyle\lim \limits_{x\to 0} \frac{ \frac{1-1-x}{1+x} }{2x} \\ &=& \displaystyle\lim \limits_{x\to 0} \frac{ \frac{ -x}{1+x} }{2x} \\ &=& \displaystyle\lim \limits_{x\to 0} \frac{ -x}{2x(1+x)} \\ &=& \displaystyle\lim \limits_{x\to 0} \frac{ -1}{2(1+x)} \quad &|\quad x = 0\\ &=& \frac{ -1}{2(1+0)} \\ &=& - \frac{1}{2} \\ \hline \end{array} \)

2. We revert the logarithm:

\(\begin{array}{|rcll|} \hline \ln \left(\displaystyle\lim \limits_{x\to 0} \left(\frac{(1+x)^{ \frac{1}{x} }}{e} \right)^{ \frac{1}{x} }\right) &=& - \frac{1}{2} \\\\ e^{\ln \left(\displaystyle\lim \limits_{x\to 0} \left(\frac{(1+x)^{ \frac{1}{x} }}{e} \right)^{ \frac{1}{x} }\right) } &=& e^{- \frac{1}{2} } \\\\ \displaystyle\lim \limits_{x\to 0} \left(\frac{(1+x)^{ \frac{1}{x} }}{e} \right)^{ \frac{1}{x} } &=& e^{- \frac{1}{2} } \\ &=& \frac{1}{ e^{ \frac{1}{2} } } \\ \mathbf{ \displaystyle\lim \limits_{x\to 0} \left(\frac{(1+x)^{ \frac{1}{x} }}{e} \right)^{ \frac{1}{x} } } &\mathbf{=}& \mathbf{\frac{1}{ \sqrt{e} } } \\ \hline \end{array}\)

Jonothan,

I am impressed with your enthusiasm in the forum but do you really think that the company will order 14.6 boxes of paper?

\(3\frac{1}{2}+1\frac{5}{7}\)

would be entered as

(3+1/2)+(1+5/7)=

(3+1/2)+(1+5/7) = 5.2142857142857143 = \(\frac{73}{14}\)

nobody can skip away from math

Source: https://www.mathsisfun.com/algebra/directly-inversely-proportional.html

y = kx 

6 = k*12

1/2 = k

The link didn't work, I will try again.

38.0

Wolfram alpha gives a step by step solution here

tan(5) = 250 / x

x*tan (5) = 250

x = 250/tan(5) = 2857.5130756878536615

Because that is what English is for.

175 reams * 1 box/ 12 reams = 175/12 = 14.6 boxes

Source: http://www.regentsprep.org/regents/math/algebra/ae7/expdecayl.htm

The formula for exponential decay (non-linear) is y = a (1 - r)^x.    Where a is the principal(initial) value of the item

                                                                                                                    r is the rate (in this case the decimal form of the percentage) at which the item decreases

                                                                                                                    x is the years over which it depreciates

Given the formula we can plug in our values:

a = $13,430

r = .041             *(4.1% /100)*

x = 7 

Our equation is : \(y= 13,430(1-.041)^x\)

Now plug 7 in for x and you get y = 13430(1-.041)^7 = $10,018.58

Originally when I did this problem I made 4.1% equal to .41 which is wrong 4.1% is equal to 0.041. My new answer is for sure correct.

Source: https://answers.yahoo.com/question/index?qid=20080701183016AAUDfF2

Sin (85) can not be found using any identities (sum and difference identities, half-angle identity, or double-angle identity). Same thing goes for cos (5). You just gotta plug it into a calculator or check the source above for some other possible ways to solve for sin(85) and cos(5)

I'm pretty sure you knew that, but this source should help you if you have any other examples: https://www.mathway.com/Precalculus.

A population of insects obeys the law of uninhibited growth. If there are 650 insects to begin with, and there are 800 after 2 days, how many will theure be after 5 days?

Thanks ClownPrinceofChaos,

Here is someone's theory of what this is all about.

This threory does not involve a lot of humour though :)     

... l feel really ashamed... trying to cram for a calculus quiz really messes up your basic math. Thanks for the correction rainstraw.

ok I looked up quadrantal angle - it is just one that is not specifically in any quadrant ie 90k degress where k is an integer.

\(sec\; \theta < 0\\ \frac{1}{cos\;\theta}\;<0\\ cos\theta <0\\ \text{This is true when }\\\theta\;\; \text{is in the second or third quad }\\\text{and when }\theta=\pm180k\;\;degrees \;\;k\in Z\)

So Metalica, your answer is correct :)

This isn't actually possible because pi is an infinite number. It doesn't end as far as we know.

2*1/2=1