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15/3+2-1+1

[15 / (3+2)] - [ 1 + 1 ] =

[15/5] - 2 =

3 - 2 =

1

Scientific notation is in the format of #.## . . . 

Only one that fits is c. 3.98*10^4.

Thank you! Btw, you came back!

Evaluate means solve or find a solution.

8.43 ~ 8

15 - 8 = 7 

A. 6.59 [~ 7].

Press the fraction button.

Courtesy of 

We need to first find the slope of the given line, 2x+4y=7.......rearranging, we have

4y = -2x + 7     divide both sides by 4

y = - 2/4x = 7/4   So, since the lines will be parallel, their slopes will be -2/4 = -1/2

And using our given point and the slope, we have

y - (-1) = -1/2(x - 3)

y + 1 =  -(1/2)x + 3/2      subtract 1 from both sides

y =  -(1/2)x + 1/2

and lots af cake (because the cake is not a lie!) 

Thank you!

and pixie dust

:l

2+2 is a mathematical expression

the answer is -8. that was an easy one, too

4 T_T c'mon dude

$${\mathtt{2}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2}} = {\mathtt{4}}$$

Yer. Well.. That's what I know of. I always just use 2 digits. 

Okay. To do front-end addition you add the 2 highest value places. And then fill the rest of the places with zeros.

$$4.11+5.90+8.02$$

We take the 2 highest value places, which are 4.1 5.9 and 8.0

$$\mathbf{4.1}0+\mathbf{5.9}0+\mathbf{8.0}2$$

Now we add these three numbers.

$${\mathtt{4.1}}{\mathtt{\,\small\textbf+\,}}{\mathtt{5.9}}{\mathtt{\,\small\textbf+\,}}{\mathtt{8}} = {\mathtt{18}}$$

How fill the other place values with zeros:

$$\mathrm{18}\textit{.00}$$

There we go.

So, do I always use the first 2 digits???

Okay. To do front-end subtraction you subtract the 2 highest value places. And then fill the rest of the places with zeros.

$$8.75-5.67$$

We take the 2 highest value places, which are 8.7 and 5.6

$$\mathbf{8.7}5-\mathbf{5.6}7$$

Now we subtract these two numbers.

$${\mathtt{8.7}}{\mathtt{\,-\,}}{\mathtt{5.6}} = {\frac{{\mathtt{31}}}{{\mathtt{10}}}} = {\mathtt{3.1}}$$

How fill the other place values with zeros:

$$\mathrm{3.1}\textit{0}$$

There we go.

Can you help me with this one?

4.11+5.90+8.02?

Okay. To do front-end addition you add the 2 highest value places. And then fill the rest of the places with zeros.

For example:

$${\mathtt{2\,310}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1\,293}}$$

We take the 2 highest value places, which are 23 and 12

$$\mathbf{23}10+\mathbf{12}93$$

Now we add these two numbers.

$${\mathtt{23}}{\mathtt{\,\small\textbf+\,}}{\mathtt{12}} = {\mathtt{35}}$$

How fill the other place values with zeros:

$$\mathrm{35}\textit{00}$$

There we go.

So, can you help me with this one?

 8.75-5.67?

Okay. To do front-end subtraction you subtract the 2 highest value places. And then fill the rest of the places with zeros.

For example:

$${\mathtt{2\,310}}{\mathtt{\,-\,}}{\mathtt{1\,293}}$$

We take the 2 highest value places, which are 23 and 12

$$\mathbf{23}10-\mathbf{12}93$$

Now we subtract these two numbers.

$${\mathtt{23}}{\mathtt{\,-\,}}{\mathtt{12}} = {\mathtt{11}}$$

How fill the other place values with zeros:

$$\mathrm{11}\textit{00}$$

There we go.

Note that the numerators are just the set of odd numbers starting with 1. The denominators are the set of perfect squares starting with 1.

The nth term is given by:

[2n - 1] / n²

Dedication.

Who's correct? CPhill or therealQuestioner?