The power, P (watts) , of a car engine is proportional to the square of its speed, s ( m / s ) . When s = 35 , P = 1580 . Work out the power (to 1 DP) when the speed is 30 m / s

P=2200 , K=35

Step-by-step explanation:

The power, P (watts), of a car engine is proportional to the square of its speed, k%28m%2Fs%29; it means if you have two different powers and speeds, you will have a proportion

P%3Ak%5E2=P%5B1%5D%3Ak%5B1%5D%5E2 or

P%2Fk%5E2=P%5B1%5D%2Fk%5B1%5D%5E2

When given:

k=35, P=2200

Work out the power P%5B1%5D (to 1DP) when the speed is k%5B1%5D=40%28m%2Fs%29

so,

P=2200

k=35

k%5B1%5D=40

and you need P%5B1%5D

2200%2F35%5E2=P%5B1%5D%2F40%5E2

2200%2F1225=P%5B1%5D%2F1600

1.79592=P%5B1%5D%2F1600

1.79592%2A1600=P%5B1%5D

P%5B1%5D=2873.5 watts

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}