A ball thrown straight up in the air travels 90 inches in the first tenth of a second. In the next tenth of a second the ball travels 85 inches. After each additional tenth of a second, the ball travels 5 inches less than it did during the preceding tenth of a second. How long will it take before the ball starts coming back down? What is the total distance that the ball has traveled once it reaches its maximum height?

855

Step-by-step explanation:

1.9 seconds is 19 tenths, so that means we add using the information given, which is that the ball travels 5 fewer inches every tenth of a second:

Time | Inches traveled

0.1     | 90

0.2    | 85

0.3    | 80

0.4    | 75

0.5    | 70

0.6    | 65

0.7    | 60

0.8    | 55

0.9    | 50

1.0     | 45

1.1      | 40

1.2     | 35

1.3     | 30

1.4     | 25

1.5     | 20

1.6     | 15

1.7     | 10

1.8     | 5

1.9     | 0

90+85+80+75+70+65+60+55+50+45+40+35+30+25+20+15+10+5+0=855

answer would be 8.64 or s = 11π/4 inches

step-by-step explanation:

s = r θ  

where s = arc length  

r = radius = 3 inches  

θ = central angle in radians = ( π /180)*165  

s = 3(π/180)*165  

= 8.64 inches

the equation of the parallel line would be y = 2x + 9

step-by-step explanation:

to start we need to find the slope of the original line. we can do that by looking at the coefficient of x, which is 2

since the slope of the original line is 2 and parallel lines have the same slope, we know the new line will also have 2 slope. now we can use that along with the point given in point-slope form to get the find equation.

y - y1 = m(x - x1)

y - 5 = 2(x + 2)

y - 5 = 2x + 4

y = 2x + 9