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You can think about negative numbers as cold numbers and positive numbers as hot numbers.

If you take away cold out of a room then the room will get hotter.  so    - - = +

$${\mathtt{\,-\,}}{\frac{{\mathtt{16}}}{{\mathtt{3}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{7}}$$

the sign always belongs to the number or letter behind it so our 2 terms here are   -16/3   and +7

It will look more normal if you swap them around

   $${\mathtt{7}}{\mathtt{\,-\,}}{\frac{{\mathtt{16}}}{{\mathtt{3}}}}$$

=   $${\mathtt{7}}{\mathtt{\,-\,}}\left({\mathtt{5}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)$$

7-5 is 2 Mmm it will have to be smaller than that because you are taking away more than 5.

=   $${\mathtt{2}}{\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{3}}}}$$

It will have to be 1 and a bit.  There are 3 thirds in a whole. ...

$$=1\frac{2}{3}$$

You will sart learning factorisation around year 8       13 years old

Permutations come in year 10 or 11                         15-16 years old

No, neither do I  MG.

Try watching this.

If you are writing in the input box you can enter

sqrt(5)     which will give you     $${\sqrt{{\mathtt{5}}}}$$             Note that this is the same as  5^(1/2)          $${{\mathtt{5}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)}$$

OR

sqrt(5,3)   which will give you     $${\sqrt[{{\mathtt{{\mathtt{3}}}}}]{{\mathtt{5}}}}$$          Note that this is the same as  5^(1/3)          $${{\mathtt{5}}}^{\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}$$

Thanks Chris

Thanks Chris

I do not understand the question.

I want to learn this!

OK...post the questions, and we'll see what we can do....!!!!

This sounds complicated but it’s just middle-school maths. Most anyone on here can answer the questions.

it was long! lol! I dont remember!

The answer isn't anything else but 3.

You need to specify which side of a triangle is 20.

Using the Law of Sines, we have

31 / sin X  = Z / sin Z

31 / (2/3)  = Z / (1/4)  simplify

93/2 = 4Z     divide both sides by 4

93/8 = Z = 11.625 

Assuming the same principal and a one year compounding period, we have....

A = (1 + .05/365)^365= 1.0512674964674626

B =  (1 + .051/12)^12 = 1.0522091760368192

C = e^.0503  = 1.0515865250168672

"B"  looks to be the best option

////////////////////////////////////

(6x^3+5x^2-3)+(3x^3-5x+6)=

We can only add/subtract "like" variables - variables having the same powers

So....rearranging, we have

[6x^3 + 3x^3] + 5x^2  - 5x + [6 - 3]  =

3x^3  + 5x^2  - 5x + 3

And that's it.....!!!

Well, BJ....don't leave us in suspense.....what answer did you get???   LOL!!!!

This is already rounded to the nearest tenth.

The answer is four (4) because: 

 + =

70k/h  = about 43.5m/h

So

Distance  / Rate = Time

180miles/43.5 miles/hour = 

about 4.137 hrs

Since four 6 x 6 squares were cut from the original, sheet, the new surface area is given by:

480 - 4(6 x 6) = 480 - 36(4)  = 480 - 144 = 336 cm^2

And the surface area of the box is given by

2[W + (W + 4)]6 = 336        where W is the width and W + 4 is the length...so we have

2[ 2W + 4]6  = 336

12[2W + 4]  = 336     divide both sides by 12

2W + 4  = 28      subtract 4 from both sides

2W  = 24           divide both sides by 2

W = 12cm    this is the width

W + 4   =  12 + 4  = 16cm      this is the length

The arc length is defined as ....  rΘ  .....where Θ  is in radians.....so... "B"  is the correct answer

Hey guys! Thank for your help!

Your suggestions and answers were not what the teacher was looking for, even though I got the exact same thing. However, I did end up getting the correct answer!

Thanks!