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harina [27]
3 years ago
7

Two window washers start at the heights shown. One is rising, the other is descending. How long does it take for the two window

washers to reach the same height? Explain.​
Mathematics
1 answer:
kvasek [131]3 years ago
8 0

Answer:

After 18.32 seconds washer men will be at the same height.

Step-by-step explanation:

Given question is incomplete; here is the complete question.

Two window washers start at the heights shown. (A: 21 ft high rising 8 in per second. The other is 50 feet high descending 11 inches per second) one is rising one is descending. How long does it take for the two window washers to reach the same height? Explain

Two window washers are moving in opposite directions.

First window washer is at 21 ft and rising with the speed of 8 inches per second.

Second window washer is at 50 feet and descending with the speed of 11 inches per second.

Since 1 feet = 12 inches

Therefore, 21 feet = 21×12 = 252 inches

Similarly, 50 feet = 50 × 12 = 600 inches

Let both the window washer takes 't' time to be at the same height.

Height of the first window washer after t seconds = (252 + 8t) feet

Height of the second window washer after t seconds = (600 - 11t) feet

Since both are at the same height after time 't',

(252 + 8t) = (600 - 11t)

8t + 11t = 600 - 252

19t = 348

t = 18.32 seconds

Therefore, after 18.32 seconds washer men will be at the same height.

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Find a positive number for which the sum of it and its reciprocal is the smallest​ (least) possible.
TiliK225 [7]

Answer:

1 is the positive number for which the sum of it and its reciprocal is the smallest.

Step-by-step explanation:

Let x be the positive number.

Then, the sum of number and its reciprocal is given by:

V(x) = x + \dfrac{1}{x}

First, we differentiate V(x) with respect to x, to get,

\frac{d(V(x))}{dx} = \frac{d(x+\frac{1}{x})}{dx} = 1-\dfrac{1}{x^2}

Equating the first derivative to zero, we get,

\frac{d(V(x))}{dx} = 0\\\\1-\dfrac{1}{x^2}= 0

Solving, we get,

x^2 = 1\\x= \pm 1

Since x is a positive number x = 1.

Again differentiation V(x), with respect to x, we get,

\frac{d^2(V(x))}{dx^2} = \dfrac{2}{x^3}

At x = 1

\frac{d^2(V(x))}{dx^2} > 0

Thus, by double derivative test minima occurs for V(x) at x = 1.

Thus, smallest possible sum of a number and its reciprocal is

V(1) = 1 + \dfrac{1}{1} = 2

Thus, 1 is the positive number for which the sum of it and its reciprocal is the smallest.

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