All Answers 358156 Answers

Hi Alan, you missed two negative signs in your definition of Inv(M).

I do not remember hearing about  this number phi before about a week ago and since then it must have been mentioned half a dozen times.  I guess I should try to slot it into my memory bank somewhere.

Good luck there.   LOL  

Use the site calc on the home page.

On the calculator on the home page here select the word Scientific on the right of the input bar and choose Equations from the pop-up menu that appears.  Then select 3 equations 3 variables.

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To get the population after 20 years simply multiply the current population by 1.12²⁰ ≈ 9.646

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Try this

"Take 2 tsp (10ml) olive oil and one (5ml) of apple vinegar mix with a little water. It will help "

A clever person once told me of this remedy for nausea - i do not know if it works.  :))

The equation resolved to 25 = a + 3 and CPhill clearly explains “ subtract 3 from both sides” showing that (a = 22)

This equation is just slightly above the 10 + 9 questions on here (just 1 number higher than the answer of 21), so it is a symptom CDD. So, you might see why I believe the disease is present.

I suppose doctors have to risk contracting the disease to help cure those who have it. CPhill demonstrated instant immunity here, but it always seems so risky.

Too bad there is not a cure for the nausea.

Yes, it means any point.  However, it usually pays to use values that make the function easy to evaluate (like -4), as long as they enable you to differentiate between the graphs you have to choose between.

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What do you mean by "How did they combine?"?  You have found the two values of x that make the original expression true.

x = -2

$${{\mathtt{3}}}^{\left({\left(-{\mathtt{2}}\right)}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\times\,}}\left(-{\mathtt{2}}\right){\mathtt{\,-\,}}{\mathtt{4}}\right)} = {\mathtt{81}}$$

$${{\mathtt{3}}}^{\left({\mathtt{2}}{\mathtt{\,\times\,}}{\left(-{\mathtt{2}}\right)}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}{\mathtt{\,\times\,}}\left(-{\mathtt{2}}\right){\mathtt{\,\small\textbf+\,}}{\mathtt{2}}\right)} = {\mathtt{81}}$$

x = -3

$${{\mathtt{3}}}^{\left({\left(-{\mathtt{3}}\right)}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\times\,}}\left(-{\mathtt{3}}\right){\mathtt{\,-\,}}{\mathtt{4}}\right)} = {\mathtt{177\,147}}$$

$${{\mathtt{3}}}^{\left({\mathtt{2}}{\mathtt{\,\times\,}}{\left(-{\mathtt{3}}\right)}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}{\mathtt{\,\times\,}}\left(-{\mathtt{3}}\right){\mathtt{\,\small\textbf+\,}}{\mathtt{2}}\right)} = {\mathtt{177\,147}}$$

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Oh I get it, I see what I missed. Thanks! I have one more question. When choosing points for the graph, it means any point? 

Approximately 27 grams in 1 mole (look at a Periodic table for a more accurate value), so 3.7*27 = 99.9 grams.

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p(-4) = 3*(3/4)^(-4+4) = 3*(3/4)⁰ = 3*1 = 3  (because any number, including 3/4, to the power 0 is 1).

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Thanks for getting back to me. Here is the equation I'm working on. It says 3 on there. 

To round to 2 decimal places look at the third decimal place.  If it is less than 5 round down, otherwise round up.  

In your case the third decimal place is 6 so we need to round the second place up. This would give us 10, so we must see this as 0 + 10, write the 0 in the second place and add a 1 to the next higher place (i.e. to the first decimal place). This results in 7.70  (you must include the last zero here as two decimal places are asked for).

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Because cos² + sin² = 1 you can turn this into a quadratic for sin by adding sin²x to both sides:

23sin²x - 2sin(x) + sin²x = cos²x + sin²x

so

24sin²x - 2sin(x) = 1

or 

24sin²x - 2sin(x) - 1 = 0

This can be solved for sinx:  

sinx = 1/4  and sinx = -1/6

Now find sin^-1x

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{1}}}{{\mathtt{4}}}}\right)} = {\mathtt{14.477\: \!512\: \!185\: \!93^{\circ}}}$$  =14.5° to nearest tenth of a degree (you will have to turn this into radians if that is what is wanted)

You can do the same for the other value.

Since it asks for all angles between 0 and 2pi that satisfy the equation, you should think about where else sinx = 1/4 etc.  (hint: look in the second quadrant when sinx is positive and in the third and fourth quadrants when sinx is negative).

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If we assume your second equation is 5x + 2y = 30 (because there are + and = signs on the same key on many keyboards) then:

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You must have written the expression incorrectly as it certainly doesn't come out as 3.

$${\mathtt{3}}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{3}}}{{\mathtt{4}}}}\right)}^{\left(-{\mathtt{4}}\right)}{\mathtt{\,\small\textbf+\,}}{\mathtt{4}} = {\frac{{\mathtt{364}}}{{\mathtt{27}}}} = {\mathtt{13.481\: \!481\: \!481\: \!481\: \!481\: \!5}}$$

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thank you so much melody and yes i meant a 0 

1^ any real number = 1

So  1^x=1

The integral of 1 is x+c

I think that is the answer 

Yes it did work! Thanks! 

$$\\log_5\;\sqrt{17}\\\\
=log_5\;17^{0.5}\\\\
=0.5\times log_5\;17\\\\
=0.5\times \frac{log17}{log5}\\\\$$

(I used the change of base law)

$${\frac{{\mathtt{0.5}}{\mathtt{\,\times\,}}{log}_{10}\left({\mathtt{17}}\right)}{{log}_{10}\left({\mathtt{5}}\right)}} = {\mathtt{0.880\: \!187\: \!213\: \!861\: \!293\: \!8}}$$

You cannot solve because there is no equal sign !

perhaps it is meant to equal zero?

If that is the case then you could use the quadratic formula to solve it :)

There is an error in your question.

You have 2 equal signs in your second equation.  

Hi Lancelot

Oscar is georgeous!  

It is so good to see you again.

That's Snoopy.  He is very friendly - just like me.  

Nauseated - are you being nausiating AGAIN!     

I wonder if there is an inoculation WE can get to protect us from your affliction?

BUT you are right CPhill does not make many mistakes and hopefully I don't either  LOL

It is good that TayJay questioned CPhill's answer.  Maybe the wording could have been a bit more tactful, but the important thing is that TayJay learned and this would not have happened if he/she did not comment on what was thought  to be an incorrect answer.