All Answers 358090 Answers

Think of this as 

$$(16z^8)^{\frac{1}{4}}\rightarrow16^{\frac{1}{4}}\times(z^8)^{\frac{1}{4}}\rightarrow2z^2$$

.

When x = 6, y = 410, so if y = m*x + 140 then 410 = m*6 + 140 

Rearrange this to find m:  m = (410 - 140)/6  or m = 45

Hence a) is the equation.

.

Pros of becoming a member:

-Great Non-Paying Job!

-Learn and Teach Math!

-Free Time Fights!

-Free Time Conversations!

-Pretty Pictures!

Best Part:

APPLYING IS FREE AND WE DONT GET PAID!!!!

IT FEELS NICE HAVING A NON PAYING JOB!!! 

Cons:

NONE!

Then grab some popcorn (and coke), and enjoy the show

$$\\\small{\text{
c. What is the largest positive integer $n$ such that 1754 and 368 }}\\
\small{\text{
leave the same remainder $r$ when divided by $n$?
}}$$

$$\small{\text{$
\begin{array}{lrcl}
(1) & 1754 &\equiv & r \pmod n\\
(2) & 368 &\equiv & r \pmod n\\
\\
\hline
\\
(1) & 1754-r &=& a\cdot n \qquad a\in Z \\
(2) & 368-r &=& c\cdot n \qquad c\in Z \qquad c\ne a\\
\\
\hline
\\
(1) & r &=& 1754- a\cdot n \\
(2) & r &=& 368- c\cdot n \\
\\
\hline
\\
(1)=(2) & 1754- a\cdot n &=& 368- c\cdot n\\
& 1386 &=& n\cdot (a-c) \qquad | \qquad n \mathrm{~is~ max, ~when ~}(a-c) = 1\\
& \mathbf{n} & \mathbf{=} & \mathbf{1386}
\\
\hline
\\
(1) & 1754 &\equiv& 368 \pmod{1386} \\
(2) & 368 &\equiv& 368 \pmod{1386} \\
\end{array}
$}}$$

$$\\\small{\text{
b. What is the largest positive integer $n$ such that 1457 and 368 }}\\
\small{\text{
leave the same remainder $r$ when divided by $n$?
}}$$

$$\small{\text{$
\begin{array}{lrcl}
(1) & 1457 &\equiv & r \pmod n\\
(2) & 368 &\equiv & r \pmod n\\
\\
\hline
\\
(1) & 1457-r &=& b\cdot n \qquad b\in Z \\
(2) & 368-r &=& c\cdot n \qquad c\in Z \qquad c\ne b\\
\\
\hline
\\
(1) & r &=& 1457- b\cdot n \\
(2) & r &=& 368- c\cdot n \\
\\
\hline
\\
(1)=(2) & 1457- b\cdot n &=& 368- c\cdot n\\
& 1089 &=& n\cdot (b-c) \qquad | \qquad n \mathrm{~is~ max, ~when ~}(b-c) = 1\\
& \mathbf{n} & \mathbf{=} & \mathbf{1089}
\\
\hline
\\
(1) & 1457 & \equiv & 368 \pmod{1089} \\
(2) & 368 & \equiv & 368 \pmod{1089} \\
\end{array}
$}}$$

@@ End of Day Wrap    Tues 14/7/15     Sydney, Australia Time   11:20pm   ♪ ♫

Hi all,

Over the past 2 days our forum users have been enlightened by Radix, Geno3141, TitaniumRome, Alan, Heureka, Sir-Emo-Chappington, asinus and fiora.  Our question askers are very fortunate to have such fine minds available to them.  Thank you.

$$\\\small{\text{
a. What is the largest positive integer $n$ such that 1457 and 1754 }}\\
\small{\text{
leave the same remainder $r$ when divided by $n$?
}}$$

$$\small{\text{$
\begin{array}{lrcl}
(1) & 1754 &\equiv & r \pmod n\\
(2) & 1457 &\equiv & r \pmod n\\
\\
\hline
\\
(1) & 1754-r &=& a\cdot n \qquad a\in Z \\
(2) & 1457-r &=& b\cdot n \qquad b\in Z \qquad b\ne a\\
\\
\hline
\\
(1) & r &=& 1754- a\cdot n \\
(2) & r &=& 1457- b\cdot n \\
\\
\hline
\\
(1)=(2) & 1754- a\cdot n &=& 1457- b\cdot n\\
& 297 &=& n\cdot (a-b) \qquad | \qquad n \mathrm{~is~ max, ~when ~}(a-b) = 1\\
& \mathbf{n} & \mathbf{=} & \mathbf{297}
\\
\hline
\\
(1) & 1754 &\equiv & 269 \pmod{297} \\
(2) & 1457 &\equiv & 269 \pmod{297}
\end{array}
$}}$$

1 mile = 1.60934 kilometres

14,000 / 1.60934 = 8700_{Miles per hour}

Thanks Alan, I wasn't thinking of anything.  I know very little physics :)

So strain does not have anything to do with breaking point then?

I mean elastic can double in length without breaking but most things break much more quickly :/

This could be made a little less "loggy" as follows:

$$\\3^{x+2}5^{x-1}=15^{2x}\\\\3^x3^25^x5^{-1}=15^{2x}\\\\(3\times5)^x\times\frac{9}{5}=15^{2x}\\\\15^x\times\frac{9}{5}=15^{2x}\\\\15^{2x-x}=\frac{9}{5}\\\\15^x=\frac{9}{5}\\\\x\times\log15=\log{\frac{9}{5}}\\\\ x=\frac{\log1.8}{\log15}$$

$${\mathtt{x}} = {\frac{{log}_{10}\left({\mathtt{1.8}}\right)}{{log}_{10}\left({\mathtt{15}}\right)}} \Rightarrow {\mathtt{x}} = {\mathtt{0.217\: \!051\: \!613\: \!246\: \!638\: \!7}}$$

heureka beat me to it!

We don't mind all your questions Mellie, they are excellent questions I wish I had more time to do them and to learn from other people's answers too.  

That is my only complaint - time is always an enemy :)

Are you understanding the answers Mellie?  You can always quiz people on their answers, if you want to, you know.

I like seeing you answer questions on the forum too.  The more that you get involved the better that it is.  

That is what I think anyway :))

solve for x: (3^(x+2))*(5^(x-1))=15^(2x)

$$\small{\text{$
\begin{array}{rcl}
3^{x+2} \cdot 5^{x-1} &=& 15^{2x}\\
3^x\cdot 3^2 \cdot 5^x \cdot 5^{-1} &=& 15^{2x}\\
3^x\cdot 5^x \cdot 3^2 \cdot 5^{-1} &=& 15^{2x}\\
( 3\cdot 5)^x \cdot 3^2 \cdot 5^{-1} &=& 15^{2x}\\
15^x \cdot \dfrac{9}{5} &=& 15^{2x}\\
15^x \cdot 1.8 &=& 15^{2x}\\
15^x \cdot 1.8 &=& 15^{x+x}\\
15^x \cdot 1.8 &=& 15^x\cdot 15^x \qquad | \qquad : 15^x\\
1.8 &=& 15^x\\
15^x &=& 1.8 \qquad | \qquad \ln{()}\\
\ln{( 15^x )} &=& \ln{ (1.8) } \\
x\cdot \ln{( 15 )} &=& \ln{ (1.8) } \\\\
x &=& \dfrac{ \ln{ (1.8) } } { \ln{( 15 )} } \\ \\
x &=& \dfrac{ 0.58778666490 } { 2.70805020110 } \\\\
\mathbf{x} & \mathbf{=} & \mathbf{0.21705161325}
\end{array}
$}}$$

√64 × 3x2y2 − 4xy × 5xy simplify

$$\\\sqrt{64}*3x^2y^2-4xy*5xy\\\\
=8*3x^2y^2-4xy*5xy\\\\
=24x^2y^2-20x^2y^2\\\\
=4x^2y^2\\\\$$

 Melody, you are right !

                        $${\mathtt{4}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}{\mathtt{\,\times\,}}{{\mathtt{y}}}^{{\mathtt{2}}}$$

Thanks from radix !

with your "Banana" and "Ends" example, that's more stating:

1 x = 2 y

where x = 2y

So yes this works out fine.

However, using 0/0 you can actually 'prove' that 1 banana = 2 bananas and then draw that to an infinite scale.

1*0 = 0

2*0 = 0

1*0 = 2*0

(1*0) / 0 = (2*0) / 0

1 = 2

$$\left({{\mathtt{3}}}^{\left({\mathtt{x}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}}\right)}\right){\mathtt{\,\times\,}}\left({{\mathtt{5}}}^{\left({\mathtt{x}}{\mathtt{\,-\,}}{\mathtt{1}}\right)}\right) = {{\mathtt{15}}}^{\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{x}}\right)}$$

We got a lot of messyness in the powers here. We can split them into more individual numbers to seperate off the constants and variables.

$${{\mathtt{3}}}^{{\mathtt{x}}}{\mathtt{\,\times\,}}{{\mathtt{3}}}^{{\mathtt{2}}}{\mathtt{\,\times\,}}{{\mathtt{5}}}^{{\mathtt{x}}}{\mathtt{\,\times\,}}{{\mathtt{5}}}^{-{\mathtt{1}}} = {{\left({{\mathtt{15}}}^{{\mathtt{2}}}\right)}}^{{\mathtt{x}}}$$

$${\mathtt{9}}{\mathtt{\,\times\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{5}}}}\right){\mathtt{\,\times\,}}{{\mathtt{5}}}^{{\mathtt{x}}}{\mathtt{\,\times\,}}{{\mathtt{3}}}^{{\mathtt{x}}} = {{\mathtt{225}}}^{{\mathtt{x}}}$$

$$\left({\frac{{\mathtt{9}}}{{\mathtt{5}}}}\right){\mathtt{\,\times\,}}{{\mathtt{5}}}^{{\mathtt{x}}}{\mathtt{\,\times\,}}{{\mathtt{3}}}^{{\mathtt{x}}} = {{\mathtt{225}}}^{{\mathtt{x}}}$$

Now we can put everything into logarithm form. Remember:

Log(a^b) = b * log(a)

Log(a*b) = log(a) + log(b)

$${log}_{10}\left({\frac{{\mathtt{9}}}{{\mathtt{5}}}}\right){\mathtt{\,\small\textbf+\,}}{\mathtt{x}}{\mathtt{\,\times\,}}{log}_{10}\left({\mathtt{5}}\right){\mathtt{\,\small\textbf+\,}}{\mathtt{x}}{\mathtt{\,\times\,}}{log}_{10}\left({\mathtt{3}}\right) = {\mathtt{x}}{\mathtt{\,\times\,}}{log}_{10}\left({\mathtt{225}}\right)$$

Let's move over all the 'x' multiples over, so we can handle them together.

$${log}_{10}\left({\frac{{\mathtt{9}}}{{\mathtt{5}}}}\right) = {\mathtt{x}}{\mathtt{\,\times\,}}{log}_{10}\left({\mathtt{225}}\right){\mathtt{\,-\,}}{\mathtt{x}}{\mathtt{\,\times\,}}{log}_{10}\left({\mathtt{5}}\right){\mathtt{\,-\,}}{\mathtt{x}}{\mathtt{\,\times\,}}{log}_{10}\left({\mathtt{3}}\right)$$

Factorise it a little, so we can now split the "x" from the rest of the formula

$${log}_{10}\left({\frac{{\mathtt{9}}}{{\mathtt{5}}}}\right) = {\mathtt{x}}{\mathtt{\,\times\,}}\left({log}_{10}\left({\mathtt{225}}\right){\mathtt{\,-\,}}{log}_{10}\left({\mathtt{5}}\right){\mathtt{\,-\,}}{log}_{10}\left({\mathtt{3}}\right)\right)$$

Simplify the logarithms and re-arrange the forumula

$${log}_{10}\left({\frac{{\mathtt{9}}}{{\mathtt{5}}}}\right) = {\mathtt{x}}{\mathtt{\,\times\,}}{log}_{10}\left({\frac{\left({\frac{{\mathtt{225}}}{{\mathtt{5}}}}\right)}{{\mathtt{3}}}}\right)$$

$${log}_{10}\left({\frac{{\mathtt{9}}}{{\mathtt{5}}}}\right) = {\mathtt{x}}{\mathtt{\,\times\,}}{log}_{10}\left({\mathtt{15}}\right)$$

$${\frac{{log}_{10}\left({\frac{{\mathtt{9}}}{{\mathtt{5}}}}\right)}{{log}_{10}\left({\mathtt{15}}\right)}} = {\mathtt{x}}$$

$${\mathtt{x}} = {\mathtt{0.217\: \!051\: \!613\: \!246\: \!638\: \!7}}$$

This is just some "brain food" as Melody said.

Anyway, your answers are perfect.

20/20

Here's your cookie:

The last three digits of 5ⁿ, when n>2, alternate between 125 (when n is odd) and 625 (when n is even).

Because 619 is odd the last three digits of 5⁶¹⁹ are 125.

That means 5⁶¹⁹/1000 = m + 0.125 where m is an integer, so 5⁶¹⁹ = 1000m + 125

i.e. The remainder on dividing 5⁶¹⁹ by 1000 is 125.

.

$$\\ \small{\text{
Several congruences are listed below and labeled with powers of 2.
}}\\
\small{\text{
Some of the congruences are true and some are false.
}}\\
\small{\text{
Find the sum of the labels for the congruences that are true.
}}\\$$

$$\small{\text{$
\begin{array}{rrcl|rcl|rcl|r}
(1) & 118 &\equiv & 25 \pmod{13} & 118 \pmod{13} &\equiv& 1 & 25\pmod{13} &\equiv& 12 & \mathrm{false}\\
(2) & 2401 &\equiv & 147 \pmod{49} & 2401 \pmod{49} &\equiv& 0 & 147\pmod{49} &\equiv& 0 & \mathrm{true}\\
(4) & 183 &\equiv & 291 \pmod{6} & 183 \pmod{6} &\equiv& 3 & 291\pmod{6} &\equiv& 3 & \mathrm{true}\\
(8) & 2701 &\equiv & 14393 \pmod{8} & 2701 \pmod{8} &\equiv& 5 & 14393\pmod{8} &\equiv& 1 & \mathrm{false}\\
(16)& 493 &\equiv & 873 \pmod{10} & 493 \pmod{10} &\equiv& 3 & 873\pmod{10} &\equiv& 3 & \mathrm{true}\\
(32)& 4113 &\equiv & 396 \pmod{9} & 4113 \pmod{9} &\equiv& 0 & 396\pmod{9} &\equiv& 0 & \mathrm{true}\\
\end{array}
$}}$$

sum of the labels for the conguences that are true: (2)+(4)+(16)+(32) = 54

Look here:

Gruß radix !

Is this your question ??  Please write !

$${\frac{{\mathtt{35}}}{{\mathtt{360}}}}{\mathtt{\,\times\,}}{\mathtt{\pi}}{\mathtt{\,\times\,}}{{\mathtt{46.15}}}^{{\mathtt{2}}} = {\mathtt{650.517\: \!264\: \!390\: \!996\: \!275\: \!9}}$$

Hi Melody,  Geno is right, Strain is a dimensionless ratio (you might be thinking of Stress which has dimensions of force per unit area).

.

thanks alan! have a nice day :)