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A rectangle is three times as long as it is wide. Determine a simplified expression that shows how the area will change if the length is decreased by 2 and the width is increased by 3.

Hello Julius,

I would like to help you work this out rather than just doing it for you.

What is the area of any rectangel?

A rectangle is three times as long as it is wide.

Let the width be w.

What is the length?

What is the area?

length is decreased by 2 and the width is increased by 3

What is the new width and length in terms of w?

What is the new area?

Now compare the old area to the new area. :)

what is 3004 in binary?

\(\begin{array}{|rclclcl|} \hline 3004 &:& 2 &=& 2\cdot 1502 &+& 0 \\ 1502 &:& 2 &=& 2\cdot 751 &+& 0 \\ 751 &:& 2 &=& 2\cdot 375 &+& 1 \\ 375 &:& 2 &=& 2\cdot 187 &+& 1 \\ 187 &:& 2 &=& 2\cdot 93 &+& 1 \\ 93 &:& 2 &=& 2\cdot 46 &+& 1 \\ 46 &:& 2 &=& 2\cdot 23 &+& 0 \\ 23 &:& 2 &=& 2\cdot 11 &+& 1 \\ 11 &:& 2 &=& 2\cdot 5 &+& 1 \\ 5 &:& 2 &=& 2\cdot 2 &+& 1 \\ 2 &:& 2 &=& 2\cdot 1 &+& 0 \\ 1 &:& 2 &=& 2\cdot 0 &+& 1 \\ \hline \end{array}\)

\(3004_{10}=101110111100_2\)

2+2 does not equal 5. Use the calculator installed in this software. 

whats the median

The median of a finite list of numbers can be found by arranging all the numbers from smallest to greatest.

If there is an odd number of numbers, the middle one is picked. For example, consider the set of numbers:

1, 3, 3, 6, 7, 8, 9

This set contains seven numbers. The median is the fourth of them, which is 6.

If there are an even number of observations, then there is no single middle value; the median is then usually defined to be the mean of the two middle values.] For example, in the data set:

1, 2, 3, 4, 5, 6, 8, 9

The median is the mean of the middle two numbers: this is (4 + 5) ÷ 2, which is 4.5

The only number I got to do math with is 4,6

Nothing more

I can't choose which part of the number or anything like that, so I'm looking for a formula which does something like that

Thank you, but modulo does not include in the "clasic" math I was thinking about, more along the lines of * / + 1 ^ 

I could do modulo manually if I had a method of flooring the number, but then I could just do it directly anyways :/

Thanks thought

How to make numbers like 4,6 to 0,6 and 5,3 to 0,3 with a mathematical formula using only "classic" math?

4,6 mod 1 = 0,6

5,3 mod 1 = 0,3

4,6 - 4 = 0,6

5,3 - 5 = 0,3

Is it actually a way to solve 0 divided by 0? I mean, you can multiply, substract and add them together. But why not divide?

Good question.   

I was going to write an answer but I decided that Google may explain it better.  You can always enter questions like this onto the google (or any other) search engine and answers will be given to you.  It is often quicker than waiting for an answer here.  We do not mind you asking here though :)

There are also some good mathematics you tubes around:

Here is one of them :)

confused

Matrix Rotation counterclockwise:

\(\begin{array}{|rcll|} \hline \begin{pmatrix} \cos(\varphi) & \sin (\varphi) \\ -\sin(\varphi) & \cos (\varphi) \\ \end{pmatrix} \stackrel{\varphi=180^{\circ}} \rightarrow \begin{pmatrix} -1 & 0 \\ 0 & -1 \\ \end{pmatrix} \\ \hline \end{array} \)

The point P becomes to P':

\(\begin{array}{|rcll|} \hline \binom{x}{y}\cdot \begin{pmatrix} -1 & 0 \\ 0 & -1 \\ \end{pmatrix} = \binom{-x}{-y} \\ \hline \end{array} \)

\(\text{Let}\ F =\binom{1}{5} \\ \text{Let}\ G =\binom{4}{-3} \)

\(\begin{array}{|rcll|} \hline F'=\binom{1}{5}\cdot \begin{pmatrix} -1 & 0 \\ 0 & -1 \\ \end{pmatrix} = \binom{-1}{-5} \\ G'=\binom{4}{-3}\cdot \begin{pmatrix} -1 & 0 \\ 0 & -1 \\ \end{pmatrix} = \binom{-4}{3} \\ \hline \end{array}\)

The answer is b.

What is 14 1/2 divided by three in fraction form

\(\begin{array}{|rcll|} \hline && \frac{14+\frac{1}{2}}{3} \\\\ &=& \frac{\frac{2\cdot14+1}{2}}{3} \\\\ &=& \frac{\frac{29}{2}}{3} \\\\ &=& \frac{29}{2\cdot 3} \\\\ &=& \frac{29}{6} \\ \hline \end{array} \)

write an equation to express the geometric sequence 10, 20, 40, 80...

\(\begin{array}{|rcll|} \hline r = \frac{20}{10} = \frac{40}{20} = \frac{80}{40} &=& 2 \\ a&=&10 \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline a_n &=& a\cdot r^{n-1} \\\\ a_n &=& 10\cdot 2^{n-1} \\ \hline \end{array}\)

check:

\(\begin{array}{|rcll|} \hline a_1 &=& 10\cdot 2^0 = 10 \\ a_2 &=& 10\cdot 2^1 = 20 \\ a_3 &=& 10\cdot 2^2 = 40 \\ a_4 &=& 10\cdot 2^3 = 80 \\ a_5 &=& 10\cdot 2^4 = 160 \\ \cdots \\ \hline \end{array}\)

I see someone likes the Percy Jackson/The Hero's of Olympus series. Good books btw. The movies suck though.

There's a calculator on the home page of this website that you can use.

To answer your question, 135.5 + 100 = 235.5

thanks

x^3 - x^2 =55   subtract 55 from both sides

x^3 - x^2  - 55   = 0

This would be difficult to solve algeraically.....here are the solutions :

http://www.wolframalpha.com/input/?i=x%5E3+-+x%5E2++-+55+++%3D+0

x ≈ 4.1672

x ≈ -1.5836 - 3.2696 i

x ≈ -1.5836 + 3.2696 i

\(\frac{(x-3)}{4}=\frac{(x-2)}{3}\)

Cross-multiply:

\(3(x-3) = 4(x-2)\)

Distribute:

3x - 9 = 4x - 8

Add 8 to both sides:

3x - 1 = 4x

Subtract 3x from both sides:

-1 = x

Actually it's undefined. You cannot divide anything into zero separate parts.