oliver is going to drive from his house to city a without stopping. oliver plans to drive at a speed of 40 miles per hour, and after driving for 3 hours, he will be 80 miles away from city a. write an equation for the function
d
(
t
)
,
d(t), representing oliver's distance from city a
t
t hours after leaving his house.

D(t)=200-40*t

Step-by-step explanation:

He was already travelling for 3 hours, so he travelled 3*40=120miles. Total distance between his house and city A is 120+80=200miles. And he is travelling 40m/h. So I think this function represent his distance from city A depending how much time he was travelling (t).

D(t)=200-40*t

D(t) = 200 - 40t

Step-by-step explanation:

Oliver lives some distance m away from city A. We know as he drives, he will get closer to city A, so his distance should decrease. so we are expecting a negative coefficient for t, let's call it r.

so we have D(t) = m - rt

we know he is driving at 40 mph, so r=40

D(t) = m - 40t

we know that after driving for 3 hours, he is 80 miles from city A.

D(3) = 80 = m - 40*3

80 = m - 120

200 = m

and so we have our equation!

D(t) = 200 - 40t

10 cm

step-by-step explanation:

if the polygon is a rectangle, then the perimeter is x+x+y+y. using the values that you gave, the perimeter is 3+3+2+2=10cm.

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}