All Answers 358095 Answers

We want the times for which (√3)sin(2t) + cos(2t) = 0.  We can rearrange this to get tan(2t) = -1/√3  (by dividing through by √3*cos(2t) and moving one term to the other side of the = sign).

Now tan(pi/6) is 1/√3, but it is negative in the second and fourth quadrants, so we must have tan(n*pi - pi/6) = -1/√3, where n is a positive integer and the angles are in radians (in degrees pi is 180 and pi/6 is 30).  

This means 2t = n*pi - pi/6 or t = n*pi/2 - pi/12.

With the calculator on the Home page here you can use the plot command.

Thanks Heureka,

Yes I know that it is true.   

I just don't understand why you even thought of it?   

I mean did something stand out to you that I am not seeing?  

It seems to have been a strange thing to have thought of.  

Hi Melody,

$$\\15848.9319246 * 10^{4.2} = 15848.9319246 * 10^{4+0.2}\\
= 15848.9319246 * 10^4*10^{0.2} \quad | \quad 10^{0.2} \approx 1.58489319246\\
= 15848.9319246 * 10^4*1.58489319246\\
\Rightarrow
=10^{0.2}*10^4* 10^4*10^{0.2}\\
=10^{8.4}$$

y ' = ½ * x^(-½) *(8x-3)^-1 + √x * [-1(8x-3)^(-2) * 8]

Do you want this to be integrated or what?

$$y'=0.5x^{-0.5}*(8x-3)^{-1}+\sqrt{x} *[-1(8x-3)^{-2}*8]\\\\

y'=0.5x^{-0.5}(8x-3)^{-1}+\sqrt{x}\; [-8(8x-3)^{-2}]\\\\$$

I've cleaned it up just a little bit but what do you want us to do with it?

Now I am curious

$${{\mathtt{10}}}^{{\mathtt{4.2}}} = {\mathtt{15\,848.931\: \!924\: \!611\: \!134\: \!852}}$$

My goodness Heureka,  what make you think of that ? ??

how to form the quadratic function into y=ax^2+bx+c  ?

$$\begin{array}{rcl}
y&=&
ax^2+bx+c\\\\
y&=&
a\left(
x^2+{b\over a}x\right)
+c\\\\
y&=&
a\left(
x^2+{b\over a}x \textcolor[rgb]{1,0,0}{
+ ({b\over 2a})^2 - ({b\over 2a})^2 }
\right)
+c\\\\
y&=&
a\left[(x+{b\over 2a})^2 - ({b\over 2a})^2
\right]
+c\\\\
y&=&a (x+{b\over 2a})^2 - {b^2\over 4a}+c\\\\
y&=&a (x+{b\over 2a})^2 + c - {b^2\over 4a}
\end{array}$$

15848.9319246 * 10^4.2 = ?

$$\boxed{15848.9319246 \approx 10^{4.2} }\\
15848.9319246 * 10^{4.2} \approx 10^{4.2} *10^{4.2}=10^{4.2+4.2}=10^{8.4}$$

15848.9319246 * 10^4.2 $$\approx$$ 10^8.4 = 251188643.151

(2/3)  / 5   =  ( 2/3 ) / (5/1)

Remember, that when we divide fractions, we "Keep"  "Change" "Flip"

"Keep" the first fraction   (2/3)

"Change" to multipliction

"Flip" the second fraction  = (1/5)

So...putting it all together, we have

(2/3) x (1/5)  =  (2/15)

what is 2 over 3 divided by 5   ?

$$\begin{array}{rcl}
\dfrac {
\frac {2}{3}
}
{
\textcolor[rgb]{1,0,0}{5}
}
=
\dfrac {
\frac {2}{3}
}
{
\textcolor[rgb]{1,0,0}{\frac {5}{1} }
}
=
\frac {2}{3}
\times
\textcolor[rgb]{1,0,0}{\frac {1}{5}}
=
\frac {2*\textcolor[rgb]{1,0,0}{1}}{3*\textcolor[rgb]{1,0,0}{5}}
=
\frac {2}{15}
\end{array}$$

Nice answer, ND........here's my slightly different approach......

Let the sum of the six numbers = S

So

S/6 = 6

And if 3 is subtracted from each of four numbers, we have...

(S -12) / 6 =

S/6 - 12/6 =

6 - 2   = 4

Let the cost be "C"

So ....

1.80C = 14      divide both sides by 1.80

C  =   14 / 1.80   ≈ $7.78

Subtract 56 from both sides

(v +3)^3 = -56    take the cubed root of both sides

v + 3  = (-56)^(1/3) = -(56)^(1/3)  = -(8*7)^(1/3) = -2(7)^(1/3)   subtract 3 from both sides

v =    -3 -2(7)^(1/3)=  -(3+ 2³√7)

38) √48x^3/√3xy^2

I have answered the question you have asked but I may not have answered the question that you intended. 

$$\frac{\sqrt{48}x^3}{\sqrt3y^2}\\\\
=\frac{\sqrt{48/3}x^3}{y^2}\\\\
=\frac{\sqrt{16}x^3}{y^2}\\\\
=\frac{4x^3}{y^2}\\\\$$

Nataszaa, why don't you post one of your teacher's answers (with the question)

CPhill is a very good mathematician, if his answer is incorrect it is most likely because you did not put brackets in and he has answered a question that is different from the one you intended.  (I may well have done the same thing)

Maybe we can teach you how to present your questions so that they will not be misinterpreted.  

15¹⁶ = 6,568,408,355,712,890,625

(over six quintillion !!!)

1.5%*500+1%*250

=0.015*500+0.01*250

$${\mathtt{0.015}}{\mathtt{\,\times\,}}{\mathtt{500}}{\mathtt{\,\small\textbf+\,}}{\mathtt{0.01}}{\mathtt{\,\times\,}}{\mathtt{250}} = {\mathtt{10}}$$

Months charges = $10

You have given an excellent answer Ninja.  

Perhaps you would like to include it in your new sample questions thread. 

Probability puzzles are never my forte. Oh well , better luck next time.

NO IT IS NOT CORRECT.  THANK YOU ANONYMOUS FOR THE CORRECTION

It seems the jury is out.  Who knows what is correct.

Since Alan is backing me I've gone back to thinking it may be alright. 

--------------------------------------------------------

5 cojoined people can be seated at a table in 4! ways.

The first couple can sit anywhere

There are 4 choices for the next 2 seats

There are 3 choices for the next 2 seats etc

5 cojoined people can be sited at a table in 4! ways.

---------------------------------------------------------

now each of the five couples can have the wife on the left or on the right that is 2 ways.

So I think that the answer is 4! * 2^5= 24*32

$${\mathtt{4}}{!}{\mathtt{\,\times\,}}{{\mathtt{2}}}^{{\mathtt{5}}} = {\mathtt{768}}$$

To find the answer, you must find the multiply 15% and $32.50.

As you see, the answer comes out as $4.875. So, round $4.875 to the nearest hunderdths place and you get $4.88.

So, you subtract $4.88 from $32.50 and the answer is $27.62.

The answer is division. If you need to find the cost of one gallon, divide $13.60 by 15 and you will get your answer.

Thanks, Ninja!

Bob grew a total of 6 inches this year. What is the average number of inches he grew per month this year?

First of all, the 3 inches Bob grew last year are not relevant to this problem (although they may be relevant to him).

There are 12 months in a year, so the average number of inches he grew per month would be the total (6) divided by the months (12)

= 0.5 inches per month.

* NinjaDevo beat me to it, sneaky Ninja! :P

The 3 inches last year isn't relavent to the question, because the question is about this year. Let's cross that right out so we don't get confused.

"Bob grew a total of 6 inches this year, which is 3 more inches than he grew last year. What is the average number of inches he grew per month this year?"

"Bob grew a total of 6 inches this year. What is the average number of inches he grew per month this year?"

There is 12 months in a year, so our answer is 6/12.

6/12 can be simplifed to 1/2

So the average growth per month is 1/2 inches.