A 18-foot ladder is leaning against a building. If the bottom of the ladder is sliding along the pavement directly away from the building at 2 ft/sec, how fast is the top of the ladder moving down when the foot of the ladder is 2 feet from the wall?

ft/sec.

Step-by-step explanation:

We have been given that a 18-foot ladder is leaning against a building. The bottom of the ladder is sliding along the pavement directly away from the building at 2 ft/sec. We are asked to find find how fast is the top of the ladder moving down when the foot of the ladder is 2 feet from the wall.

First of all, we will draw a graph to represent the given information as shown in the attachment.

Now, we will use Pythagoras theorem as:

Now, we will find derivative with respect to time.

Let us find value of x, when y is 2 using Pythagoras theorem.

Upon substituting our given values in derivative function, we will get:

Therefore, the top of the ladder is moving down at a rate of feet per second.

a; x = 0.5

step-by-step explanation:

-10x = -5

x = 0.5

step by step resolutions

1) n+n+1+n+2=51

3n+3=51

3n=48

n=16

the three numbers are 16, 17 and 18.

2) x+15=4x

3x=15

x=5

radha is now 5 years old

3) the two digits are x and y.

y=3x

(10x+y)+(10y+x)=88

y=3x

11x+11y=88

y=3x

x+y=8

x+3x=8

x=2

the original number can be either 28 or 82.

4) n+(n+2)+(n+4)=48

3n+6=48

3n=42

n=14

the numbers are 14, 16 and 18.

5) y=x+3

10y+x+10x+y=143

y=x+3

11x+11y=143

y=x+3

x+y=13

x+x+3=13

2x=10

x=5

the original number can be 57 or 75.

hope this !