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

yes

step-by-step explanation:

the correct answer is

c. 4

step-by-step explanation:

the given polynomial is

the degree of the polynomial is the highest exponent which is 4.

the -4 is called the leading coefficient.

this polynomial has only one term so it is a monomial.

the correct answer is option c.