The relationship between altitude and the boiling point of a liquid is linear. At an altitude of 8100 ft, the liquid boils at 203.66 degrees F. At an altitude of 4700 ft, the liquid boils at 208.42 degrees F. Write an equation giving the boiling point b of the liquid, in degrees Fahrenheit, in terms of altitude a, in feet. What is the boiling point of the liquid at 2600 ft?

9514 1404 393

Step-by-step explanation:

We are given two points: (a, b) = (8100, 203.66) and (4700, 208.42). We can use these in the 2-point form of the equation for a line

  y = (y2 -y1)/(x2 -x1)(x -x1) +y1 . . . . . . line through (x1, y1) and (x2, y2)

Using the given values, we have ...

  b = (208.42 -203.66)/(4700 -8100)(a -8100) +203.66

  b = 4.76/-3400(a -8100) +203.66

  b = -0.0014a +215 . . . . . simplify

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At 2600 ft, the boiling point is ...

  b = -0.0014(2600) +215 = 211.36 °F

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}

answer: 40392

step-by-step explanation:

10098 x 4 = 40392