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I think that this is a link ot download the web2.0 calc

I think that this is it.

3x^2+1=4x

$$\\3x^2-4x+1=0\\\\
3x^2-3x-1x+1=0\\\\
3x(x-1)-1(x-1)=0\\\\
(3x-1)(x-1)=0\\\\
x=1/3 \qquad or \qquad x=1$$

2x + y = 7          ...(1)

  x + y = 4          ...(2)

Subtract (2) from (1) term by term

2x - x + y - y = 7 - 4

Simplify

x = 3

Put this back into (2

3 + y = 4

Subtract 3 from both sides

y = 1

.

4x^2-x=0

Please note: If the question was not so specific you would never so this one this way!

$$\\4x^2-x=0\\\\
x^2-\frac{1}{4}x=0\\\\
x^2-\frac{1}{4}x+(\frac{1}{2}\times\frac{1}{4})^2=0+(\frac{1}{2}\times\frac{1}{4})^2\\\\
x^2-\frac{1}{4}x+(\frac{1}{8})^2=0+(\frac{1}{8})^2\\\\
x^2-\frac{1}{4}x+\frac{1}{64}=\frac{1}{64}\\\\
x^2-\frac{1}{4}x+\frac{1}{64}=\frac{1}{64}\\\\
(x-\frac{1}{8})^2=\frac{1}{64}\\\\
x-\frac{1}{8}=\pm\frac{1}{8}\\\\
x=\frac{1}{8}+\frac{1}{8}\qquad or \qquad x=-\frac{1}{8}+\frac{1}{8}\\\\
x=\frac{1}{4}\qquad or \qquad x=0 \\\\$$

Where did the -(-1) come from Chris?

I like the rest of the explanation.  

Oh dear yes I did mean 5.  Thanks Chris.   

I ate so much for dinner that I cannot fit in any sweets (super) anyway  - so there.  

You can't!  The calculator here doesn't yet have a summation function.

.

x^2 = 5(x + 100)       distribute the 5 across the terms in the parentheses

x^2 = 5x + 500         subtract the terms on the right from both sides

x^2 - 5x - 500 = 0     factor

(x - 25) (x + 20) = 0   set each factor to 0

x = 25    or   x = -20

Този въпрос изглежда непълна.

(This question seems incomplete.)

Yeah...the steps look OK.

Remember that the "range" for the tangent inverse (i.e., the values returned by the function) are from -pi/2 to pi/2, exclusively. And -pi/2 and 3pi/2 are really the same coterminal angles.

I always think of it as 

gradient = gradient

Then all line formulas are the same. :)

4 + 4 - 4/4 =

8 - 1 =

7

Are you here ?

Melody meant to say that x = 5 rather than 11.....

Just for that, young lady, you'll be sent to your room with NO supper !!!!

I think this person wants to write an equation of a line using the slope and a point. We will actually have the same answer as Melody....it will just look a little different!!!

So using the point-slope form, we have

y - y₁ = m(x - x₁)       where m is the slope and (x₁, y₁) is the given point   ..... so we have

y - 2 = 3(x - 1)          add 2 to each side and distribute the "3" across the terms in the parentheses

y = 3x - 3 - 2            simplify

y = 3x - 5                 

Note.....putting our given values of 2 and 1 in for x and y will make this equation true.....so we know we have done things correctly !!!

I like to explain it in this manner:

Suppose we have a  30 gallon bucket half full of water that has a hole in the bottom.. And water leaks out of the bucket at one gallon per minute  →   (-1) gallon per minute.

And suppose that time going forward is positive, and time going backward is negative.

So, five minutes from now, the bucket will contain (-1)*(5) = -5 gallons less than it does now.

But, 10 minutes ago, the bucket contaned (-1)*(-10) = 10 gallons more than it does now.

So, in the first case, a negative times a positive is a negative. And in the second case, a negative times a  negative is a positive !!!!

HI Melody, thanks it is a good video.

Have a look through this post and see if you can't work it out by yourself.

$$3=\frac{y-2}{x-1}$$

then simplify :)

y=-2log3 all over 5log3-7log2?

Do you want

$$\\y=\frac{-2log3}{5log3-7log2}\qquad?\\\\
y=\frac{log3^{-2}}{log3^5-log2^7}\\\\
y=\frac{log3^{-2}}{log\frac{3^5}{2^7}}\\\\
$This does not seem to be helping much$$$

$${\mathtt{\,-\,}}{\frac{{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{log3}}}{\left({\mathtt{5}}{\mathtt{\,\times\,}}{\mathtt{log3}}{\mathtt{\,-\,}}{\mathtt{7}}{\mathtt{\,\times\,}}{\mathtt{log2}}\right)}} = {\mathtt{\,-\,}}{\frac{{\mathtt{2}}{\mathtt{\,\times\,}}{log}_{10}\left({\mathtt{3}}\right)}{{\mathtt{\,-\,}}{\mathtt{7}}{\mathtt{\,\times\,}}{log}_{10}\left({\mathtt{2}}\right){\mathtt{\,\small\textbf+\,}}{\mathtt{5}}{\mathtt{\,\times\,}}{log}_{10}\left({\mathtt{3}}\right)}}$$      that didn't help either

I did it on my casio calc and got -3.427640726

This is an estimate.

My goodness that is one energetic monotreme!!!   Someone has been telling you porkies.  

I seriouly hope that you are not Australian.   LOL

5455/(16*365.25)  km/day

$${\frac{{\mathtt{5\,455}}}{\left({\mathtt{16}}{\mathtt{\,\times\,}}{\mathtt{365.25}}\right)}} = {\frac{{\mathtt{5\,455}}}{{\mathtt{5\,844}}}} = {\mathtt{0.933\: \!436\: \!002\: \!737\: \!850\: \!8}}$$

I don't know how to insert -2log3 divided by (5log3-7log2) in the calculator, to get an answer

$$y=\frac{-2log3}{5log3}-7log2$$

is that your question?

If it is then you can just cancel out the log3 s

If it is not then you need to introduce brackets and post under this one.  

You can use the web2c calc on the home page.  Enter it as   (2+1/2)*(3/4)