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Square root 49 and 64 to get the number that you square to get it.

$${\mathtt{5\,185}}{\mathtt{\,\times\,}}{\mathtt{0.7}} = {\frac{{\mathtt{7\,259}}}{{\mathtt{2}}}} = {\mathtt{3\,629.5}}$$

£34.29

I'll do the middle one, you can do the other two.

f(x) = 3x - 4,

so

f(1.5) = 3(1.5) - 4 = 4.5 - 4 = 0.5.

?

It comes from the word implied.

There is an implied relationship  between the two variables.

It may or may not be possible (or desirable) to convert it into an explicit relationship.

$${\left({{\mathtt{b}}}^{{\mathtt{5}}}\right)}^{{\mathtt{2}}} = {\left({\mathtt{b}}{\mathtt{\,\times\,}}{\mathtt{b}}{\mathtt{\,\times\,}}{\mathtt{b}}{\mathtt{\,\times\,}}{\mathtt{b}}{\mathtt{\,\times\,}}{\mathtt{b}}\right)}^{{\mathtt{2}}}$$ = $${\mathtt{b}}{\mathtt{\,\times\,}}{\mathtt{b}}{\mathtt{\,\times\,}}{\mathtt{b}}{\mathtt{\,\times\,}}{\mathtt{b}}{\mathtt{\,\times\,}}{\mathtt{b}}{\mathtt{\,\times\,}}{\mathtt{b}}{\mathtt{\,\times\,}}{\mathtt{b}}{\mathtt{\,\times\,}}{\mathtt{b}}{\mathtt{\,\times\,}}{\mathtt{b}}{\mathtt{\,\times\,}}{\mathtt{b}}$$ = $${{\mathtt{b}}}^{{\mathtt{10}}}$$

$${\left({{\mathtt{b}}}^{{\mathtt{5}}}\right)}^{{\mathtt{2}}} = {{\mathtt{b}}}^{{\mathtt{10}}}$$

$${{\mathtt{b}}}^{{\mathtt{5}}}{\mathtt{\,\times\,}}{{\mathtt{b}}}^{{\mathtt{2}}} = {{\mathtt{b}}}^{{\mathtt{7}}}$$

x - 70 = 760

x = 760 + 70         =>      x = 830

Kn = Ko * (1+p/100)^n

Ko = 50 000 000 $        p = 20 %         n = 2  years

Kn =$${\mathtt{50\,000\,000}}{\mathtt{\,\times\,}}{{\mathtt{1.2}}}^{{\mathtt{2}}} = {\mathtt{72\,000\,000}}$$  $

You will have  72 000 000 $ .

Greetings radix !

ty

your my role model

Simply divide the numerator by the denominator and it will be in decimal form

For example if i wish to convert 7/8 to decimal form i will perform the following: Seven divided by eight = .875

What number do we need to multiply by 10 to get 20???

That number will be your answer.......

@@ End of Day Wrap  Thurs 11/6/15   Sydney, Australia Time   12:30am  (Yes it is Fri)   ♪ ♫

Hi everyone,

Today our fabulous answers were provided by CPhill Badinage, Bertie, TitaniumRome, Headingnorth, Alan Heureka Radix and DarkBlaze347.   Thanks you  

Thanks Alan   

Here's a simple example comparison.

Non-implicit ( y = f(x) )

y = x + 1          

dy/dx = 1

Implicit ( y = f(x,y) )

y = x*y + 1

dy/dx = y*1 + x*dy/dx

(1 - x)dy/dx = y

dy/dx = y/(1 - x)

But (1 - x)y = 1, or y = 1/(1 - x), so, substituting this in the above, we get dy/dx = 1/(1-x)²

In this case, from y = 1/(1-x) we could have written directly;  dy/dx = -(1-x)^-2*(-1)  or  dy/dx = 1/(1-x)², the same as above.  

However, we can't always rearrange y = f(x,y) to get y = f(x) explicitly, so sometimes we are forced to use implicit differentiation.

.

.

determine the interest on $5000 invested at 12% per annum compound interest for 3 years, with the interest compounded a)monthly?

I.  12% per annum compound interest for 3 years:

$$\small{\text{$
\begin{array}{rcl}
\$5000\cdot \left(1+\dfrac{12}{100}\right)^3\\\\
&=&\$5000\cdot 1.12^3\\\\
&=& \$7024.64
\end{array}
$}}$$

II.  12% per annum with the interest compounded monthly

$$\small{\text{$
\begin{array}{rcl}
\$5000\cdot \left(1+\dfrac{12}{100}\cdot \dfrac{1}{12}\right)^{36}\\\\
&=&\$5000\cdot 1.01^{36}\\\\
&=& \$7153.84
\end{array}
$}}$$

well im pretty sure that you open your eyes or move to a corner and try to hide that your sleeping

Thanks guys   

Find the horizontal asymptote of  f(x) = 2 ((x+4)(5 x-1))/((8-x)(7 x + 2))

$$\small{\text{$
\lim \limits_{x \rightarrow \infty } 2\cdot \dfrac{(x+4)(5x-1)}{(8-x)(7x+2)}
=\lim \limits_{x \rightarrow \infty } 2\cdot \dfrac{5x^2+19x-4}{-7x^2+54x+16}
=\lim \limits_{x \rightarrow \infty } 2\cdot \dfrac{ \dfrac{5x^2}{x^2}+\dfrac{19x}{x^2}-\dfrac{4}{x^2}}{ \dfrac{-7x^2}{x^2}+\dfrac{54x}{x^2}+\dfrac{16}{x^2}}
=\lim \limits_{x \rightarrow \infty } 2\cdot \dfrac{ 5 + \dfrac{19}{x}-\dfrac{4}{x^2}}{ -7 + \dfrac{54}{x}+\dfrac{16}{x^2}}
= 2\cdot \dfrac{ 5 }{ -7 }=-\dfrac{10}{7} $}}$$

$$\rm{horizontal~ Asymptote:~} y =-\dfrac{10}{7} = -1.42857142857$$