All Answers 358098 Answers

$$\begin{array}{rll}
\frac{3}{10}&=&\frac{?}{20}\\\\
\frac{3*2}{10*2}&=&\frac{?}{20}\\\\
\frac{6}{20}&=&\frac{?}{20}\\\\
6&=&?\\\\
?&=&6\\\\

\end{array}$$

$$\begin{array}{rll}
5(m+1)&=&7(m-1)\\
5m+5&=&7m-7\\
5m+5-5&=&7m-7-5\\
5m&=&7m-12\\
5m-7m&=&7m-12-7m\\
-2m&=&-12\\
\frac{-2m}{-2}&=&\frac{-12}{-2}\\
m&=&6\\
\end{array}$$

Do you mean on the web2 calculator?

inverse trig is the same as arc trig

so if you want inverse tan just use atan

5. The sin is < 0 in quadrants 3 and 4. And the cos < 0 in quadrants 2 and 3. But the sin and the cos  would have to have the same sign for the tangent to be positive. Thus, theta is a 3rd quadrant angle. And since the tan(theta)= 1 at 45 degrees, this is a 45 degree angle in quadrant 3. And the cos of 45 degrees  = 1/√2, or √2/2, if you prefer. But, since the cos is negative in this quadrant, then cos(theta) = - 1/√2 = -√2/2.

6.  

2 cos x = -1   divide both sides by 2

cos x = -1/2    ....    this occurs at (2/3)pi in the interval [0, pi)  and at -(2/3)pi in the interval [-pi, 0]. The second answer may be a little hard to see. Note that cos x = -1/2 at (4/3)pi. But, subtracting 2*pi from this gives us -(2/3)pi.......

The arcsin of (9/16) isn't 55.8. The arccos of (9/16) will give you that answer. Go to the onsite calculator. Hit the "2nd" button and you will see the "acos" button. Hit this, key in 9/16, hit "Enter" or the "=" key and you will get your result.

What rate do you have and I will show you how to work out the effective  annual rate. 

for Q 5 ... so waht is the answer of the question !!! 

for Q 6 ... the answer in the model answer is   +and -  2pi/3

I'm wondering if the phrasing of this question should be "at least two."

Consider the following pentagon in which the vertices represent 5 people at a gathering........call them A,B, C,D, E.

Now, suppose that all the people there are total strangers to one another. Then, it's clear that there are more than two people who have the same number of friends at the gathering. In fact, there are 5 people who have the same number of friends.......namely ......none!!!

But, if the phrase includes"at least two," then it's true. Suppose we have some polygon with a "large number" of alphabetical vertices - call the number "n" -, and let each vertex represent a "person."  Now, connect every vertex with every other vertex and let these connections represent the associations between people. In this instance everyone knows the same number of people - (n-1). Now, erase any one of the connected lines. Then, the two vertexes (people) that were connected by this line now know the same number of people (n-2). And if we continued in this fashion of erasing lines, there would always be at least two vertices that had the same valence (valence is the number of lines meeting at one vertex). And thus, there would always be at least two people who knew the same number of people!! For instance, suppose we erased AB  and AC. Then B and C would know the same number of people - in this case, everybody else except "A." Or, if we erased AB, BC and AC, then A, B, and C would know the same number of people - (n-3).

I realize that this isn't a "formal" proof, but I think it demonstrates the idea. Perhaps someone on the site could give a more "formal" one??? I'm sure there's a simpler way, I just don't happen to know it....!!!

5m+5=7m-7=6 6 is the answer

Multiply the numerator by 2 since 10*2 = 20 --> 3*2 = 6 --> 6/20.

OK! Here is my proof!

Lemma: $${{\mathtt{e}}}^{\left({\mathtt{\pi}}{\mathtt{\,\times\,}}{i}\right)} = -{\mathtt{1}}$$

Statement                                          Reason

1.$${{\mathtt{e}}}^{\left({\mathtt{\pi}}{\mathtt{\,\times\,}}{i}\right)} = {\mathtt{x}}$$                                   Given

2.$${{\mathtt{e}}}^{\left({\mathtt{k}}{\mathtt{\,\times\,}}{i}\right)} = \underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{k}}\right)}{\mathtt{\,\small\textbf+\,}}{i}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{k}}\right)}$$              Euler's Identity

3.$${{\mathtt{e}}}^{\left({\mathtt{\pi}}{\mathtt{\,\times\,}}{i}\right)} = \underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{\pi}}\right)}{\mathtt{\,\small\textbf+\,}}{i}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{\pi}}\right)}$$            From 1. and 2.

4.$${{\mathtt{e}}}^{\left({\mathtt{\pi}}{\mathtt{\,\times\,}}{i}\right)} = {\mathtt{\,-\,}}{\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{i}{\mathtt{\,\times\,}}{\mathtt{0}}$$                 Simplify

5.$${{\mathtt{e}}}^{\left({\mathtt{\pi}}{\mathtt{\,\times\,}}{i}\right)} = -{\mathtt{1}}$$                                Simplify

Q.E.D.

Main Proof:

Statement                                                   Reason

1.$${{i}}^{{i}} = {\mathtt{x}}$$                                                    Given

2. $${ln}{\left({{i}}^{{i}}\right)} = {ln}{\left({\mathtt{x}}\right)}$$                                    Take log of both sides

3.$${i}{\mathtt{\,\times\,}}{ln}{\left({i}\right)} = {ln}{\left({\mathtt{x}}\right)}$$                                 Power rule of logarithms

4. $${i} = {\sqrt{-{\mathtt{1}}}}$$                                             Given

5.$${i} = {\left(-{\mathtt{1}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)}$$                                         Algebra

6.$${i}{\mathtt{\,\times\,}}{ln}{\left({\left(-{\mathtt{1}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)}\right)} = {ln}{\left({\mathtt{x}}\right)}$$                      From 5. and 3.

7. $${i}{\mathtt{\,\times\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right){\mathtt{\,\times\,}}{ln}{\left(-{\mathtt{1}}\right)} = {ln}{\left({\mathtt{x}}\right)}$$                Power rule of logarithms

8. $${ln}{\left({\mathtt{x}}\right)} = {\mathtt{y}}$$ if and only if $${{\mathtt{e}}}^{{\mathtt{y}}} = {\mathtt{x}}$$              Definition of ln

9.$${i}{\mathtt{\,\times\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right){\mathtt{\,\times\,}}\left({\mathtt{\pi}}{\mathtt{\,\times\,}}{i}\right) = {ln}{\left({\mathtt{x}}\right)}$$                 Use the lemma

10. $$\left({\frac{\left({\mathtt{\pi}}{\mathtt{\,\times\,}}{i}{\mathtt{\,\times\,}}{i}\right)}{{\mathtt{2}}}}\right) = {ln}{\left({\mathtt{x}}\right)}$$                            Algebra

11. $$\left({\frac{\left(\left(-{\mathtt{1}}\right){\mathtt{\,\times\,}}{\mathtt{\pi}}\right)}{{\mathtt{2}}}}\right) = {ln}{\left({\mathtt{x}}\right)}$$                           Algebra

12. $$\left({\frac{-{\mathtt{\pi}}}{{\mathtt{2}}}}\right) = {ln}{\left({\mathtt{x}}\right)}$$                                      Algebra

13. $${{\mathtt{e}}}^{\left({\frac{-{\mathtt{\pi}}}{{\mathtt{2}}}}\right)} = {\mathtt{x}}$$                                          Take antilog of both sides and use step 8.

14.$${\frac{{\mathtt{1}}}{\left({{\mathtt{e}}}^{\left({\frac{{\mathtt{\pi}}}{{\mathtt{2}}}}\right)}\right)}} = {\mathtt{x}}$$                                          Algebra

15. $${\frac{{\mathtt{1}}}{\left({{\sqrt{{\mathtt{e}}}}}^{\,{\mathtt{\pi}}}\right)}} = {\mathtt{x}}$$                                            Algebra

16. $$\left({\frac{{\mathtt{1}}}{\left({{\sqrt{{\mathtt{e}}}}}^{\,{\mathtt{\pi}}}\right)}}\right){\mathtt{\,\times\,}}\left({\frac{{{\sqrt{{\mathtt{e}}}}}^{\,\left({\mathtt{4}}-{\mathtt{\pi}}\right)}}{{{\sqrt{{\mathtt{e}}}}}^{\,\left({\mathtt{4}}-{\mathtt{\pi}}\right)}}}\right) = {\mathtt{x}}$$              Rationalize the denominator

17. $${\frac{\left({{\sqrt{{\mathtt{e}}}}}^{\,\left({\mathtt{4}}-{\mathtt{\pi}}\right)}\right)}{\left({{\sqrt{{\mathtt{e}}}}}^{\,{\mathtt{4}}}\right)}}$$                                             Algebra

18. $${\frac{{{\sqrt{{\mathtt{e}}}}}^{\,\left({\mathtt{4}}-{\mathtt{\pi}}\right)}}{{{\mathtt{e}}}^{{\mathtt{2}}}}} = {\mathtt{0.207\: \!879\: \!576\: \!350\: \!761\: \!9}}$$   Algebra

And this is a real number.

Q.E.D.

Thanks to you all for submitting your proofs and also, please respond to tell me of any flaws in my proof. I will try to get back to WebCalc 2.0 with another puzzle proof question soon!

130000*0.0057 = 741.

11.045

First, we can take out the 11 leaving 0.045.

Now, if you multiply 0.045 by 1000, you get 45...so 45/1000

Along with the 11, we have $$11\frac{45}{1000}$$.

We can reduce this by dividing both the numerator and denominator of 45/1000 by 5.

$$\mathbf{11\frac{9}{200}.}$$

2/k - 18 = 1/k

-18 = 1/k - 2/k

-18 = -1/k

-18k = -1

k = -1/-18

k = 1/18.

For whole numbers: divide by 1, for example: 6=6/1, -3=-3/1.

To convert from mixed fraction to improper fraction: multiply the whole number by the denominator and add the answer to the numerator

Example: 5 1/2 = (5*2+1)/2=11/2

Definitions:

• Proper Fractions Numerator < Denominator
  Proper fractions have the nominator part smaller than the denominator part,
  for example , or  .
• Improper Fractions Numerator > Denominator or Numerator = Denominator,
  Improper fractions have the nominator part greater or equal to the denominator part,
  for exampleor .
• Mixed Fractions
  Mixed fractions have a whole number plus a fraction, for example 2or 123 .

tan theta = sin theta / cos theta

sin theta / cos theta = 1

sin theta = cos theta

If sin theta < 0, then cos theta must be equal to sin theta in order for tan theta to equal 1.

2 cos x = -1

cos x = -1/2

We would be looking at the 2nd quadrant...x has to be 120 degrees or 2pi/3.

1000 is the same as 10*10*10 or simply 10 to the third power, 10³.

sqrt(180)

= sqrt(9*4*5)

=sqrt(9)*sqrt(4*5)

= 3*2*sqrt(5)

= 6*sqrt(5)

= $$\mathbf{6\sqrt{5}.}$$

x^x when x=6, is the same as 6x6x6x6x6x6=46,656

thank you for your help Alan but at the end i solved it with the S│V│T method and again, thank you so much for helping me ☺☻☺☻☺☻

Let taxi speed be v km per hour, so truck speed is 0.75v kph.

Truck travels for 3 hours, so distance travelled by truck in that time = 0.75v*3 = 9v/4 km

Taxi travels for 2 hours, so distance travelled by taxi in that time = 2v km

These must total 360 km so:   2v + 9v/4 = 360 or 17v/4 = 360

v = 4*360/17 = 84.71 kph

So truck speed = 0.75*4*360/17 = 63.53 kph

Just to add to this, here's how you turn 14 and 2/3 into an improper fraction:

First multiply 14 times 3, add 2 to that number, and put that number over 3.

$$14\frac{2}{3}=\frac{(14\times 3)+2}{3} = \frac{(42)+2}{3}=\frac{44}{3}$$

wouldnt it easier to solve it this way $${\frac{\left({\frac{{\mathtt{14}}}{{\mathtt{1}}}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{2}}}{{\mathtt{3}}}}\right)}{{\mathtt{4}}}} = {\frac{{\mathtt{11}}}{{\mathtt{3}}}} = {\mathtt{3.666\: \!666\: \!666\: \!666\: \!666\: \!7}}$$ with the calculater?

$$Turn $14\frac{2}{3}$ into an improper fraction = $\frac{44}{3}$\\\\
Divide this by 4 to get $\frac{11}{3}$ or $3\frac{2}{3}$$$

Great job Melody my proof is probably a bit less elegant and is also longer. You got the right answer though.