Solve (t - 3)2 = 6. The arrow is at a height of 48 ft after approximately _ s
and after _s

The arrow is at a height of 48 ft after approximately 0.55 s  and after 5.45 s

Step-by-step explanation:

The following information is missing:

The height of an arrow shot upward can be given by the formula s = v0*t - 16*t², where v0 is the initial velocity and t is time.How long does it take for the arrow to reach a height of 48 ft if it has an initial velocity of 96 ft/s?

If the arrow is at a height of 48 ft and its initial velocity is 96 ft/s, then:

48 = 96*t - 16*t²

16*t² - 96*t + 48 = 0

16*(t² - 6*t + 3) = 0

t² - 6*t + 3 = 0

t² - 6*t + 3 + 6 = 0 + 6

t² - 6*t + 9 = 6

(t - 3)² = 6

t - 3 = √6

t - 3 = 2.45; t = 2.45 + 3; t = 5.45

or

t - 3 = -2.45; t = -2.45 + 3; t = 0.55

3/4 ÷ 3/4

step-by-step explanation:

basically divide any proper fraction with itself to fit the criteria.

so if we are using 3/4, then:

3/4 ÷ 3/4 = 1

so 1 is the quotient.

and the numerators for both these fractions is 3.   and if we divide the numerators:

3 ÷ 3 = 1

we see that quotient is equal to our original quotient as well.

answer: 42

use a calculator to find that 845/73 = 11.575

the whole part of the answer is 11. multiply this by 73 to get 11*73 = 803.

subtract that result from 845 to get 845 - 803 = 42