Plane A leaves the Calgary airport at a bearing of 70 degrees and travels a distance of 420km. Plane B leaves the Calgary airport at a bearing of 280 degrees and travels a distance of 340km. How far apart are the planes to the nearest kilometre.

16650 km

Step-by-step explanation:

If plane A leaves from Calgary airport at a nearing of 70 degree, it would end up on the first quadrant. Since we are not given exact bearing here we would assume both planes begin from the first quadrant, they move from North. Example plane A moves North 70 degrees.

Plane B will move North 280 degrees and end up in the fourth quadrant (because each quadrant is 90 degrees so we would have 90+90+90+10=280)

Distance apart = 70+80=150 degrees( since all four quadrants =360 degrees)

1 degree= 111km (approximate metric equivalent of degrees)

Therefore distance apart = 150 degrees ×111km = 16650km

step-by-step explanation:

30% of $420 is 0.30($420) = $126 and $420 + $126 = $546.

or, 130% of $420 is 1.3($420) = $546.

notice that this is not the $600 original list price with which we started.

why doesn't this work?

the method does not work because 30% of $420 is not the same amount as the 30% of $600. it may seem obvious when stated this way, but it is a very common error made on problems of this type. we cannot just use the same percent and get back to the original amount because we are taking that percent of a different value.

to work problems of this type correctly, it is generally best to write an equation for the situation and then solve.

example: since (100% – 30%)(original price) = (sale price)

70% of the   (original price) is $420.

0.7x   = 420

equation image indicator

x = 600

the original cost was $600.

we may also solve this problem using the two-step method. working the problem in two steps requires us to use more algebra skills. we have to know that x = 1x (identity property of multiplicationtext annotation indicator) and that 1x – 0.3x = 0.7x (subtracting like terms). then we solve the problem as follows:

(original price) – 30%(original price) = sale price

x – 0.3x = 420

0.7x = $420

equation image indicator

x = 600

the original cost was $600.  

note that with both of the above methods we do obtain the correct original amount.

example: after a 5% pay raise, hermione is earning $22,680 per year. what was she earning before the pay raise?

105% of (original salary) = current salary

1.05x = 22,680

equation image indicator

x = 21,600

hermione earned $21,600 per year before the pay raise.

this is an example so maybe you will get how to at least do it!

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step-by-step explanation: