We can calculate the depth ddd of snow, in centimeters, that accumulates in Harper's yard during the first hhh hours of a snowstorm using the equation d=5hd=5hd, equals, 5, h. How many hours does it take for 111 centimeter of snow to accumulate in Harper's yard? hours How many centimeters of snow accumulate per hour? centimeters

a) Number of hours it takes 1 centimetre of snow to form in Harper's yard = (1/5) hour = 0.20 hour

b) Centimetres of snow that accumulate per hour = 5 cm

Step-by-step explanation:

Complete Question

We can calculate the depth d of snow, in centimeters, that accumulates in Harper's yard during the first h hours of a snowstorm using the equation d=5h.

a) How many hours does it take for 1 centimeter of snow to accumulate in Harper's yard? hours

b) How many centimeters of snow accumulate per hour? centimeters

Solution

The depth of snow, d, in centimetres that accumulates in Harper's yard in h hours is given d = 5h

a) Number of hours it takes 1 centimetre of snow to form in Harper's yard.

d = 5h

d = 1 cm

h = ?

1 = 5h

h = (1/5) = 0.20 hour

b) Centimetres of snow that accumulate per hour.

d = 5h

In 1 hour, h = 1 hour

d = ?

d = 5 × 1 = 5 cm

Hope this Helps!!!

1/5 hours

5 centimeters

Step-by-step explanation:

no oblique asymptote

step-by-step explanation:

to find the oblique asymptote of f(x)

here numerator has degree 1 and denomintor degree 2

degree of numerator < degree of denominator

hence there is no oblique asymptote

when the degree of the denominator > that of the numerator, we saw that the value in the denominator got so much bigger, so quickly,   giving us a horizontal asymptote of the x-axis. or we can write

f(x) = a polynomial of x + a proper fraction.

in this case only we have oblique asymptotes but not when numerator has less degree.