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what is 50=x/6+8 re-arrange

x/6 =50 - 8

x/6 =42 multiply both sides by 6

x = 252

Thank you. The question was a bit above my level.

Simply replace t by 4 in your formula:

N(4) = 10*3^4 → 10*81 → 810

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Hi GerardWay

Yes, Heureka and Alan are VERY impressive !!         

It's a recursivie function.

\(fib (n) = fib(n-2) + fib(n-1) \text{ iff } n \geq 2 \\ fib(n) = 1 \text{ iff } n = 0 \text{ } \vee n = 1\)

Well maybe but I am not too sure because he was pretty square.

The calc on the home page does do that.   

Just press the equal sign a second time and it will display in scientific notation.    

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The calc on the forum won't do this though  

What is I+1=x I-1=x I=??????

huh??  

What is the question ???

Hi Vest4R

The question is

sin(theta + pi/4) = 1/2

now (theta + pi/4) is an angle so I could substitute    alpha

\(sin\;\alpha=\frac{1}{2}\)

A lot of people just memorize that sin30 degrees=1/2 but I memorize the triangle that it come from. 

I draw an equilateral triangle with sides 2 units and cut it in half.

The right angle triangle thus formed will be this:

Note that 30 degrees is   pi/6 radians

so

\( sin \frac{\pi}{6}=\frac{1}{2}\\ \alpha=\frac{\pi}{6}\\ so\\ \theta+\frac{\pi}{4}=\frac{\pi}{6}\\\)

Does that fully answer your question?    

If you ask more questions you may be advised to private message me this address cause otherwise I may not see it. :)

what is the fibonacci sequence

Fibonacci numbers: F(n) = F(n-1) + F(n-2) with F(0) = 0 and F(1) = 1.

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368, 75025, 121393, 196418, 317811, 514229, 832040, 1346269, 2178309, 3524578, 5702887, 9227465, 14930352, 24157817, 39088169 ...

I.e. You start with two 1's and each subsequent term is the sum of the previous two terms.

1

1

2

3

5

8

13

...

1

1+0=1

1+1=2

1+2=3

2+3=5

5+3=8

8+5=13

...

Sorry, without mistakes:

2) Express in the form a + bi:

\(\ln(3+4i)\)

Formula:

\(\begin{array}{|rcll|} \hline \ln(a+i\cdot b)=\ln(\sqrt{a^2+b^2}) + i\cdot \arctan(\frac{b}{a}) \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \ln(3+4i) \qquad a=3 \qquad b = 4 \\ \ln(3+4i) &=& \ln(\sqrt{3^2+4^2}) + i\cdot \arctan(\frac{4}{3}) \\ \ln(3+4i) &=& \ln(5) + i\cdot 0.92729521800\dots \\ \ln(3+4i) &=& 1.609437912434100\dots + i\cdot 0.92729521800161\dots \\ \hline \end{array}\)

Thank you for your responce melody,

this may be rather simple, but would you be able to explain show how you went from the start of the question to: \(sin\frac{\pi}{6}= \frac{1}{2}\)

Regards,

Almost right EP, but 15/10 is 1.5.  

No.  The integral is with respect to r not F.

.

Here's a way of doing the first part of 4):

The second integral can be done in similar fashion (though there are higher powers to deal with).  You should find the result is \(\frac{5\pi}{256}a^8\)

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565.45 g

there are 75+44 students in grade 4

i hope it's useful

In a Cavendish experiment that first measured gravity, 

a 0.730 kg lead ball was placed 0.230 m away from a 158 kg lead ball.

How much gravitational force did that create?

Fg=G(m1m2)/r^2

G=6.67*10^-11

Let m₁ = 0.730  kg

Let m₂ = 158  kg

Let r = 0.230  m

Let G = \(6.67\cdot 10^{-11} \ N\cdot \frac{m^2}{kg^2}\)

\(\begin{array}{|rcll|} \hline F_g &=& G\ \frac{m_1\cdot m_2}{r^2} \\ F_g &=& 6.67\cdot 10^{-11} \ N\cdot \frac{m^2}{kg^2}\ \cdot \frac{0.730\ kg\cdot 158\ kg}{0.230^2\ m^2} \\ F_g &=& 6.67\cdot 10^{-11} \cdot \frac{0.730\cdot 158}{0.230^2} \ N\cdot \frac{m^2}{kg^2}\ \cdot \frac{ kg^2 }{m^2} \\ F_g &=& 6.67\cdot 10^{-11} \cdot \frac{0.730\cdot 158}{0.230^2} \ N \\ F_g &=& 6.67\cdot 10^{-11} \cdot \frac{115.34}{0.0529} \ N \\ F_g &=& 6.67\cdot 10^{-11} \cdot 2180.34026465 \ N \\ F_g &=& 6.67\cdot 10^{-11} \cdot 2.18034026465\cdot 10^3 \ N \\ F_g &=& 6.67\cdot 2.18034026465\cdot 10^{-11} \cdot 10^3 \ N \\ F_g &=& 6.67\cdot 2.18034026465\cdot 10^{-8} \ N \\ F_g &=& 14.5428695652 \cdot 10^{-8} \ N \\ F_g &=& 1.45428695652\cdot 10^1 \cdot 10^{-8} \ N \\ F_g &=& 1.45428695652\cdot 10^{-7} \ N \\ F_g &=& 1.45428695652\cdot 10^{-1}\cdot 10^{-6} \ N \\ F_g &=& 0.145428695652\cdot 10^{-6} \ N \\ \mathbf{F_g} & \mathbf{=} & \mathbf{0.145428695652\ \mu N } \\ \hline \end{array}\)

Start by adding 2 to both sides.

Then divide both sides by -5

the current through a resistor of unknown value is 0.025 A when it is connected to a 10.0- V source.

What is the value of the resistor

Ohm's Law : V = I*R

\(\begin{array}{|rcll|} \hline R&=& \frac{V}{I} \\ R&=& \frac{10.0 V}{0.025\ A} \\ R&=& 400\ \frac{V}{A} \\ R&=& 400\ \Omega \\ \hline \end{array} \)

The value of the resistor is 400 \(\Omega\)

sin(theta + pi/4) = 1/2

\(sin(\theta + pi/4) = 1/2\)

Looking at just the acute angle.

\(sin\frac{\pi}{6}=\frac{1}{2}\)

sine is positive in the first and second quads so ...

\(\theta+\frac{\pi}{4}=\frac{\pi}{6},\quad \frac{5\pi}{6},\quad \frac{13\pi}{6}\\ \theta+\frac{3\pi}{12}=\frac{2\pi}{12},\quad \frac{10\pi}{12},\quad \frac{26\pi}{12}\\ \theta=\frac{2\pi}{12}-\frac{3\pi}{12},\quad \frac{10\pi}{12}-\frac{3\pi}{12},\quad \frac{26\pi}{12}-\frac{3\pi}{12}\\ \theta=\qquad \frac{-1\pi}{12},\qquad \frac{7\pi}{12},\qquad \frac{23\pi}{12}\\ but\;\;0\le\theta\le2\pi\\ so\\ \theta= \frac{7\pi}{12},\qquad \frac{23\pi}{12}\\ \)