Sir-Emo-Chappington

You use the inverse-sine function. Common notation is either "asin(x)" or "sin^-1(x)"
This returns the angle that has a sine of x.

asin(0.836) = 56.72° (To 2 d.p.)

Think of it like this:

The price you are paying ($20) is 85% of the original price (100% - 15%).

So, $20 = Original Price * 85%

Re-arrange and you get:

Original Price = $20 / 85%

= $23.53 (Rounded up to the nearest cent)

Once again I give incorrect answers because I skim over a detail >_<'

I shall make it my task to read things in their entirety from now on!

For whatever the number, divide it by Pi to get how many times bigger than Pi it is (Hence giving as a multiple of Pi).

tan(x) = 11/r

cos(x) = r/s

sin(x) = ?

Using the equation:

$$Tan(x) = Sin(x) / Cos(x)$$

Therefore:

11/r = Sin(x) / (r/s)

11/r = (s * Sin(x)) / r

Since both sides are divided by "r", let's multiply it out.

11 = s * Sin(x)

We want Sin(x) on it's own, so divide both sides by "s".

11 / s = Sin(x)

Sin(x) = 11 / s

41_mph * t_hours = Total distance

54_mph * 3.7_hours = Total distance

Therefore: 41_mph * t_hours = 54_mph * 3.7_hours

Divide both sides by 41 to make "t" the subject of the equation.

t = (54 * 3.7) / 41

= 199.8 / 41

= 4.8731707317073171

= 4.9_hours (to 1 d.p.)

I completely skimmed over the detail the first corridor is only 8m wide -_- I thought they were the same width. Alan has the right idea.

14a - 93 = 49 - 57a

First let's re-arrange the equation so all "a"s are on one side.

14a + 57a - 93 = 49

14a + 57a = 49 + 93

Simplifying it down

71a = 142

Divide both sides by 71 so it's just "a" as the subject.

a = 142 / 71

a = 2

Amount deposited on monday = x

The amounts he has left by the end of each day:

Monday: x

Tuesday: x + 700

Thursday: (x+700) - 300 - 200 = x + 200

Friday: (x + 200) / 2

So (x + 200)/2 = 1400 remaining in the account

Multiply both sides by 2

x + 200 = 2800

Subtract both sides by 200

x = 2600

The amount deposited on monday was 2600

For question 2, consider this:

The girder must rotate a full 90 degrees (Or pi/4 if you like your radians), so at one point or another it's going to be at 45° to the corridor walls. Seeing as we ignore width of the beam we can forget about it 'catching' on the walls when it rotates.

So we have a nice situation of all we need is the corridor corner to support a three-quarter circle of diameter 27 metres. This is a radius of a 27/2 = 13.5 metres. So the corridor needs to be 13.5 metres wide.

If my explanation is a bit confusing, just say and I will try to mock up some diagrams to better explain it.