All Answers 359166 Answers

Well...if the second term was 4, the formula would be nⁿ   where n is an integer > 0

Using the change of base rule.....

x = log(10,000,000)/log(2.512) ≈ 17.5

43/60*100  percent

So, elizabeth.....you joined up just for this???  That's the definition of "Get A Life"

254 x 12 = 3048

Thanks Chris  :)

Very impressive Heureka :)

Since you have had no answer it is probably safe to assume that i am not the only one confused by your question.  :)

A division and a fraction sign are the same thing.

On the forum you can put it in using the Math Formula input or by using the LaTex Formula input.

The Math Formula one is easier to use.

$${\frac{{\mathtt{5}}}{{\mathtt{6}}}}$$          this is 5/6 put in the Math formula

 $$\frac{5}{6}$$         and this is  \frac{5}{6}   entered into the LaTex formula input box

Is that any help?

The language is Maylay:

How do you do division in conventional form?


 The question was posted 988 days ago, near the bottom of the pyramid. 

1298cosx=2770sinx+325.8

$$\begin{array}{rcl}
1298*cos(x) &=& 2770*sin(x)+325.8\\
\underbrace{1298}_{x_p}*\underbrace{cos(x)}_{n_x}
-\underbrace{2770}_{y_p}*\underbrace{sin(x)}_{n_y} &=& \underbrace{325.8}_d \qquad \text{line: } \vec{p}*\vec{n}=d \quad \vec{p}=(x_p,y_p)=(1298,-2770) \\
\tan{(\alpha)} &=& \frac{y_p}{x_p} = -\frac{2770}{1298}\\\\
cos(x-\alpha) &=& \frac{d}{ \sqrt{x_p^2+y_p^2} } \\\\
sin(x-\alpha) &=& \frac{ \sqrt{x_p^2+y_p^2-d^2}}{\sqrt{x_p^2+y_p^2} }\\\\
tan(x-\alpha) &=& \pm \sqrt{\frac{ x_p^2+y_p^2 }{d^2} -1 )} \\\\
x_{1,2} -\alpha &=& tan^{-1} { \left( \pm \sqrt{\frac{ x_p^2+y_p^2 }{d^2} -1 \right) } }\\\\
x_{1,2} &=& \alpha + tan^{-1} { \left( \pm \sqrt{\frac{ x_p^2+y_p^2 }{d^2} -1 \right) } }\\\\
x_{1,2} &=& tan^{-1} { \left( \frac{y_p}{x_p} \right) } + tan^{-1} { \left( \pm \sqrt{\frac{ x_p^2+y_p^2 }{d^2} -1 \right) } }\\\\
x_{1,2} &=& tan^{-1} { \left( \frac{-2770}{1298} \right) } + tan^{-1} { \left( \pm \sqrt{\frac{ (1298)^2+(-2770)^2 }{(325.8)^2} -1 \right) } }\\\\
x_{1,2} &=& ( -64.8925847874\ensurement{^{\circ}} \pm n*\pi) +(\pm 83.8861674476\ensurement{^{\circ}} \pm n*\pi)\\\\
x_1 &=& -64.8925847874\ensurement{^{\circ}} + 83.8861674476\ensurement{^{\circ}} \pm n*2\pi\\\\
x_2 &=& -64.8925847874\ensurement{^{\circ}} - 83.8861674476\ensurement{^{\circ}} \pm n*2\pi\\\\
x_1 &=& 18.9935826602\ensurement{^{\circ}} \pm n*2\pi\\\\
x_2 &=& -148.778752235\ensurement{^{\circ}} \pm n*2\pi
\end{array}$$

Thanks Heurka, a great answer as always.  

There is a calculator on the home page - why don't you consult it?

You cannot divide by zero.  The answer is undefined.

This is what Princess kitty says about it

something^16

Yes it is where y=0

The roots or zeros are the the x intercepts .  :)

In general, the roots of a function, f(x), say, are the values of x that make f(x) zero.

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See if the following graph helps

.

Good job by Melody and heureka....points for both........maybe some of my detective work is rubbing off on Melody ..... LOL!!!

What are the coordinate Points on the graph and the Intersection

x-Intersection = 0

y-Intersection = 0

We can model this with the following inequality

50x + 160 ≤ 1000      where x is the number of boxes....so, subtracting 160 from each side, we have

50x ≤ 840                  divide both sides by 50

x ≤ 16.8 boxes     ....but since he can't bring up partial boxes, round down to 16

Huh? What language are you writing in?

What are the Points on the graph and the Intersection?

x- Intersection ( 0.5 ,  0 )

y- Intersection ( 0 ,  -1 )

6/(2/3)

$${\frac{{\mathtt{6}}}{\left({\frac{{\mathtt{2}}}{{\mathtt{3}}}}\right)}} = {\mathtt{9}}$$

Like that - is that what you mean?

@heureka Please.... What are the coordinate Points on the graph and the Intersection? Please write it out. Thanks.

@heureka Please.... What are the Points on the graph and the Intersection? Please write it out. Thanks