A 13 foot ladder is leaning against a brick wall. The bottom of the ladder is pulled away at the rate of 3ft/sec. How fast is the top of the ladder sliding down the bricks when the bottom is 5 feet from the wall?

the top of the ladder is sliding down the bricks at - 1.25 ft/s

Step-by-step explanation:

The ladder forms a right angle triangle with the ground. The length of the ladder represents the hypotenuse.

Let x represent the distance from the top of the ladder to the ground(opposite side)

Let y represent the distance from the foot of the ladder to the base of the wall(adjacent side)

The bottom of the ladder is pulled away at the rate of 3ft/sec. This means that y is increasing at the rate of 3ft/sec. Therefore,

dy/dt = 3 ft/s

The rate at which x is reducing would be

dx/dt

Applying Pythagoras theorem which is expressed as

Hypotenuse² = opposite side² + adjacent side², it becomes

x² + y² = 13²- - - - - - - -1

Differentiating with respect to time, it becomes

2xdx/dt + 2ydy/dt = 0

2xdx/dt = - 2ydy/dt

Dividing through by 2x, it becomes

dx/dt = - y/x ×dy/dt- - - - - - - - - - 2

Substituting y = 5 into equation 1, it becomes

x² + 25 = 169

x² = 169 - 25 = 144

x = √144 = 12

Substituting x = 12, dy/dt = 3 and y = 5 into equation 2, it becomes

dx/dt = - 5/12 × 3

dx/dt = - 5/4 = - 1.25 ft/s

t = 3

step-by-step explanation:

put t = 3 in this expression :

(3²-1)(3²-2)(3² - 3)(3² - 4) = 8×7×6×5 = 56×30 = 1680

7

step-by-step explanation:

i'm assuming you need to figure out how many campers go into each tent.

i got the answer by:

24+4=28

28/4=7