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That is a really good answer Thejamesmachine   

Right method zac, but 0.001m is 1mm not 10mm.

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That is an excellent answer Zac  

Here is a video to reinforce what Zac has told you   

Melody, You use the "Input" line below the graph.  Just type in y=x^2-4*x and press Enter.  Similarly for x=4.

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That is neat Alan  

This is a perfect square

$$x^2+2bx+b^2=(x+b)^2$$

Can you work out the answer to your own question now?

n = 51

$${\left(-{\mathtt{1}}\right)}^{{\mathtt{51}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}{\mathtt{\,\times\,}}{\mathtt{51}} = {\mathtt{50}}$$

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I don't know how to do this problems BUT are the b***s identical or different from one another?

That is an excellent answer anon 

I love those crocodile tears Civonamzuk!     

didig    Stop giving people negative points - it will come back to haunt you!   

You would do the exponent first. They come after brackets but before multiplication, division etc. (PEDMAS)

Hello anonymous,

3 x 3 x 3 x 3 x 3 = 3^5 = 243

gruß :- )

how many decimeters is a meter .  1 m = 10 dm

We can use the Law of Cosines to solve this........so we have....

(1900^2  - 1875^2 - 2125^2) / (-2(1875)*2125)  = cosA

cos^-1 ( (1900^2  - 1875^2 - 2125^2) / (-2(1875)*2125)) =  about 56.3°

Well it depends what you are asked to find . . .

The intercept/gradient form is y = mx + b, where m = gradient and b = the y-intercept

so y = 2*(x + 2)     ------>    y = 2x + 4

so m = 2

the y-intercept = 4

To find the x-intercept we can make y = 0 . . .

0 = 2x + 4

2x = -4

x = -4/2 = -2

the x-intercept = -2

If you are asked to graph it, then start on a point you know (the y-intercept), which has the coordinates of (0,4). For your other point you could either use your x-intercept or another point based on your gradient. 

We know that m = rise/run. 2 can be written as 2/1, where for every positive unit moved on the run (x), it moves 2 on the rise (y)

So starting at (0,4) and moving accross 1 on the x we rise 2 on the y. So (0+1,4+2) = (1,6)

So you could graph it using the points (0,4) and (1,6), drawing a continuous line though them and continuing on indefinately in both directions.

Sorry if I am being too long winded/confusing. Here is it graphed so you can visualise it.

The quadratic formula is ususally an option but here I would probably do this:

$$8y^2 + 26y + 15$$   

( Note that 8 does not go easily into the other terms, making factorising by straight up removing the 8 out of the question. Instead we can look at the product of 8 and 15, with the intention of splitting 25 up into 2 terms which can then be factorised. What 2 numbers add to make 26 and multiply to make (8*15) 120? 6 and 20)

$$8y^2 + 6y + 20y + 15$$

( Now we factorise in pairs)

$$2y(4y + 3) + 5(4y + 3)$$

$$(2y + 5)(4y + 3)$$

YAY :)

Just to check.....

$$(2y + 5)(4y + 3)$$

$$(2y\times{4y}) + (2y\times{3}) + (5\times{4y}) + (5\times{3})$$  (FOIL)

$$8y^2 + 6y + 20y + 15$$

$$8y^2 + 26y + 15$$

We can look at this in terms of a rate. . .

1m : 1000 years ...... Divide both sides by 1000

0.001 m : 1 year ........ change units

1 mm : 365 days ......... divide by 365

0.0027397mm (roughly) : 1 day ........ * by 2.5

0.0068493 mm (roughly) : 2.5 days

(corrected)

Try this:

3,1

1,2

3,3

1,4

You can use the Change-of-Base formula:

If you want the new base to be 'b' and the number that you want to find the log of 'x':

log_b(x)  =  log₁₀(x) / log₁₀(b)     or     log_b(x)  =  ln(x) / ln(b)

X = 18

Plug in 2 for "Y" and you get 6. Subtract 6 from 24 to get "X".

x+3y=24

y=2

X=?

This is a substitution question.

So by substituting in [y = 2] . . .

x + 3*(2) = 24

x + 6 = 24

x = 18