All Answers 358179 Answers

look here

there is a calc on the home page.

7/2 + 2109* (3218*32)

4*1.6*10^(9) = 6.4*10^9

Here are a couple more options.

$$\\ln(x^2)=1\\
2lnx=1\\
lnx=0.5\\
x=e^{0.5}\\$$

I have a further question Heureka. 

$$\!\ \varphi(6) = 2. \qquad 1 \text{ and } 5 \\\\
6 = 2 \cdot 3\\
\!\ \varphi(6)}=6\cdot (1-\frac{1}{2})( 1-\frac{1}{3} )}=2$$

Now I do get why phi(6)=2

But what is it about from 6=2.3

What is the relvance of the brackets.  How can this be used on other examples?

Thank you Heureka,  that explains it beautifully 

$$\\V=\frac{4}{3}\pi*r^3\\
estimated\;r\;=21cm\qquad max\;r\;=21.05cm\\
$biggest error = max volume - estimated volume$\\
max\; error=\frac{4}{3}\pi*21.05^3-\frac{4}{3}\pi*21^3\\\\
max\; error=\frac{4}{3}\pi*(21.05^3-21^3)\\\\$$

$${\frac{{\mathtt{4}}}{{\mathtt{3}}}}{\mathtt{\,\times\,}}{\mathtt{\pi}}{\mathtt{\,\times\,}}\left({{\mathtt{21.05}}}^{{\mathtt{3}}}{\mathtt{\,-\,}}{{\mathtt{21}}}^{{\mathtt{3}}}\right) = {\mathtt{277.748\: \!730\: \!102\: \!649\: \!218\: \!5}}$$

max possible error approx  278  cm^3

almost :)

$$x=\pm\sqrt{2}$$

I am sorry but the question is wrong and so is your answer silver  

f(x)=-x^2-3x+1

find    f(-4)

f(-4)=-(-4)^2-3*-4+1 = -16+12+1=-3

$$\\tan(x+50)=cot(x )\\
tan(x+50)=tan(90-x )\\
x+50=90-x\\
2x=40\\
x=20\\$$

I expect there are other answers but that is one anyway.

I suspect that you have not interpreted this as intended silver27.

This is why people must learn to use brackets properly .   

maybe, if j = sqrt(-1)  then

$$\\e^{-2j}\\\\
=cos(-2)+isin(-2)\qquad $in radians$\\\\
\approx \;-0.4161-0.9093j$$

Could a mathematician please check this.   

Hello Melody,

$$\\\mathbf{De{finition:}}\\
\begin{text}
L{et} $ n \ge 1$ be an integer. Then we de{fine} the \\
\textit{Euler phi function} $\phi$ by\\
$\phi(n)=$ the number of positive integers \\
less than $n$ that are relatively prime to $n$
\end{text}$$

$$\\\mathbf{Relatively Prime:}\\
$
\begin{text}
Describes two numbers for which\\
the only common factor is 1. \\
In other words, relatively prime numbers have\\
a greatest common factor $(gcf)$ of $1$. \\
For example, $6$ and $35$ are relatively prime $(gcf = 1)$.\\
The numers $6$ and $8$ are not relatively prime $(gcf = 2)$.
\end{text}$$

$$\\\mathbf{Example\ 1: n = 6 }
$ and n is not a prime number $\\
\samll{\text{
$
\begin{array}{rr}
\textcolor[rgb]{1,0,0}{1}& gcf(6,1) = \textcolor[rgb]{1,0,0}{1}\\
2& gcf(6,2) = 2\\
3& gcf(6,3) = 3\\
4& gcf(6,4) = 2\\
\textcolor[rgb]{1,0,0}{5}& gcf(6,5) = \textcolor[rgb]{1,0,0}{1}\\
6& gcf(6,6) = 6\\
\end{array}
$
}}\\
6 \text{ has } \textcolor[rgb]{0,0,1}{2} \text{ relative primes } 1 \text{ and } 5 \text { see the red color. So } \phi(6) = \textcolor[rgb]{0,0,1}{2}$$

$$\\\mathbf{Example\ 2: n = 9 }
$ and n is not a prime number $\\
\samll{\text{
$
\begin{array}{rr}
\textcolor[rgb]{1,0,0}{1}& gcf(9,1) = \textcolor[rgb]{1,0,0}{1}\\
\textcolor[rgb]{1,0,0}{2}& gcf(9,2) = \textcolor[rgb]{1,0,0}{1}\\
3& gcf(9,3) = 3\\
\textcolor[rgb]{1,0,0}{4}& gcf(9,4) = \textcolor[rgb]{1,0,0}{1}\\
\textcolor[rgb]{1,0,0}{5}& gcf(9,5) = \textcolor[rgb]{1,0,0}{1}\\
6& gcf(9,6) = 3\\
\textcolor[rgb]{1,0,0}{7}& gcf(9,7) = \textcolor[rgb]{1,0,0}{1}\\
\textcolor[rgb]{1,0,0}{8}& gcf(9,8) = \textcolor[rgb]{1,0,0}{1}\\
9& gcf(9,9) = 9\\
\end{array}
$
}}\\
9 \text{ has } \textcolor[rgb]{0,0,1}{6} \text{ relative primes } 1,2,4,5,7,8 \text { see the red color. So } \phi(9) = \textcolor[rgb]{0,0,1}{6}$$

$$\\\mathbf{Example\ 3: n = 7 }
$ and n is a prime number $\\
\samll{\text{
$
\begin{array}{rr}
\textcolor[rgb]{1,0,0}{1}& gcf(7,1) = \textcolor[rgb]{1,0,0}{1}\\
\textcolor[rgb]{1,0,0}{2}& gcf(7,2) = \textcolor[rgb]{1,0,0}{1}\\
\textcolor[rgb]{1,0,0}{3}& gcf(7,3) = \textcolor[rgb]{1,0,0}{1}\\
\textcolor[rgb]{1,0,0}{4}& gcf(7,4) = \textcolor[rgb]{1,0,0}{1}\\
\textcolor[rgb]{1,0,0}{5}& gcf(7,5) = \textcolor[rgb]{1,0,0}{1}\\
\textcolor[rgb]{1,0,0}{6}& gcf(7,6) = \textcolor[rgb]{1,0,0}{1}\\
7& gcf(7,7) = 7\\
\end{array}
$
}}\\
\small{
7 \text{ has } 7-1=\textcolor[rgb]{0,0,1}{6} \text{ relative primes } 1 \text{ until } 6 \text { see the red color. So } \phi(7) = \textcolor[rgb]{0,0,1}{6}.}\\
\text{ So } \phi \text{ of a prime number } \phi(p) = p-1$$

Look in the sticky topics !!!!!

Thanks Chris, that is high praise. 

just eneter this in the calc and youll get the answer!

no i dont think so

One way is known as the "carpenter's trick."

To find out if two walls are at right angles to each other, the carpenter marks a point on one of the walls 3 units from where the surfaces join. On the other wall, at the same height, he marks a point 4 units from where the walls join.Then, he measures the distance between the points. If this = 5 units, the walls are at right angles.

This "works" brcause of the Pythagorean "Triple,"   3 - 4  - 5

Original force on charge 1 from the others f = K(q1*q2/d² + q1*q3/(√2*d)² + q1*q4/d²) where K is a constant (I've assumed d1 and d3 are on opposite corners).

(Strictly, we should add these as vectors, but that's an unnecessary complication here.)

If all the q terms are multiplied by 2 and everything else remains the same it is easy to see that a factor of 4 increases each term, and hence the force on charge 1 quadruples.  The same argument applies to all the other charges, so SevenUp is correct.

.

Here's one possibility......{it's the graph of the cardioid,  r = 8 - 8sin(Θ) }

If the 50° refers to the angle between the ladder and the ground then you have a right angled triangle for which you know the angle and hypotenuse and want to find the "opposite", so you have:

sin(50°) = opposite/hypotenuse or opposite = hypotenuse*sin(50°) so opposite = 6.5*sin(50°) m

. 

We can't tell......how many pupils can one enumerator survey in one day???

a).  Substitute 1 and 2 into the expression for R(n) to get two simultaneous equations for h and k:

n = 1:       h^1/2/k = 4.8           (working on millions of dollars)

n = 2:       h¹/k  = 5.76  from this we have h = 5.76k.  Put this into the first expression above to get:

(5.76k)^1/2/k = 4.8    or 5.76^1/2/k^1/2 = 4.8.   Multiply both sides by k^1/2 and divide both sides by 4.8 to get:

k^1/2 = 5.76^1/2/4.8.   Square both sides:  k = 5.76/4.8²

$${\mathtt{k}} = {\frac{{\mathtt{5.76}}}{{{\mathtt{4.8}}}^{{\mathtt{2}}}}} \Rightarrow {\mathtt{k}} = {\mathtt{0.25}}$$

Hence from the n = 2: equation above we have h = 5.76*0.25 or 

$${\mathtt{h}} = {\mathtt{5.76}}{\mathtt{\,\times\,}}{\mathtt{0.25}} \Rightarrow {\mathtt{h}} = {\mathtt{1.44}}$$

Remember that the result of using these values in R(n) = h^n/2/k will be in millions of dollars.

For the sum of R(n) to n, notice that it can be written as:

sum(n) = (a + a² + a³ + ... + aⁿ)/k  where a = h^1/2

The numerator is a geometric progression giving the result of sum(n) = a*(aⁿ-1)/[(a-1)k] 

or sum(n) = (h^1/2)*(h^n/2-1)/[(h^1/2-1)k}

See if you can now have a go at part b).

.

Let a₁  = the first term....we have....the sum of the 1st 6 terms  = 42....therefore...

42 = (6/2)[2a₁ + 5d]

42 = 3[2a₁ + 5d]

14 = 2a₁ + 5d  →   a₁ = [14 - 5d]/2 

And a₃₀ = 3(a₁₀)   .... therefore.....

3(a₁₀) = a₁₀ + 20d

2(a₁₀) = 20d

a₁₀ = 10d

So we have

a₁₀ = (a₁ + 9d)  ... and by substitution.....

10d = ([14- 5d]/2 + 9d)

20d = 14 - 5d + 18d

20d = 14 + 13d

7d = 14

d = 2

Therefore....

a₁ = [14 - 5(2)] / 2  = [14 - 10] / 2 =  4/2 = 2

And

a₁₃ = [2 + 2(12)] =  26