a ladder 5 meters long is leaning against a wall. the base of the ladder is sliding away from the wall at a rate of 1 meter per second. how fast is the top of the ladder sliding down the wall at the instant when the base in meters from the wall?

Complete question :

a ladder 5 meters long is leaning against a wall. the base of the ladder is sliding away from the wall at a rate of 1 meter per second. how fast is the top of the ladder sliding down the wall at the instant when the base is 4 meters from the wall?

-4/3 m/s

Step-by-step explanation:

Using Pythagoras :

a² = x² + c²

5² = x² + y² - - - (1)

The horizontal distance x with respect to time t = 1m/sec

dx/dt = 1m/sec

To obtain the vertical height 'y' with respect to time t when x = 4

5² = x² + y²

Wen x = 4

5² = 4² + y²

25 = 16 + y²

y =√(25 - 16)

y = 3

Differentiate (1) with respect to t

x² + y² = 25 - - (1)

2x * dx /dt + 2y dy/dt = 0

dx/dt = 1

x = 4.

y = 3

2(4)(1) + 2(3) * dy/dt = 0

8 + 6dy/dt = 0

6 dy/dt = - 8

dy/dt = - 8/6

dy/dt = - 4/3

ratio is 2 : 9

the answer is three

13.5 : 3

step-by-step explanation:

part a. the differences are decreasing, so we know the sequence is not arithmetic. the ratios are constant (=3/4), so we know the sequence is geometric.

part b. from part a, we know the common ratio is 3/4, so each term is 3/4 of the one before. the initial term is the number of gallons at h=1.

g[h] = (3/4)·g[h-1]

g[1] = 384 . . . . initial value required for a recursive definition

part c. the general formula for a geometric sequence is

a[n] = a[1]·r^(n-1)

we have a[1] = 384, r = 3/4, and the term number given by h. the formula that follows is

g[h] = 384·(3/4)^(h-1)

part d. filling in the value and doing the arithmetic, we have

g[12] = 384·(3/4)^11 ≈ 384·0.042235136 ≈ 16.2183 . . gallons

part e. each gallon is 4 quarts, so

16.2183 gal × 4 qt/gal = 64.8732 qt