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3 years ago

a minute hand of a clock is 22 feet long.Find the distance traveled by the the tip of the minute hand in one hour.

Mathematics

1 answer:

Allushta [10]3 years ago

7 0

Answer:

138.23 feet

Step-by-step explanation:

The minutes and seconds hands of a clock rotates in a circular manner no matter the shape of the clock. Therefore the distance covered by the minute hand of the clock is calculated by the circumference of a circle:

C = 2πr

C = 2 * $\frac{22}{7}$ * 22

C = 138.23 feet

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$$\frac{50-2 w^{2}}{3 w^{2}+9 w-30} \cdot \frac{w^{2}+5 w-14}{6 w-30}=-\frac{w+7}{9}$

Solution:

Given expression:

$$\frac{50-2 w^{2}}{3 w^{2}+9 w-30} \cdot \frac{w^{2}+5 w-14}{6 w-30}$

To solve the given expression:

First simplify: $\frac{50-2 w^{2}}{3 w^{2}+9 w-30}$

$$\frac{50-2 w^{2}}{3 w^{2}+9 w-30}=-\frac{2(w+5)(w-5)}{3(w-2)(w+5)}$

Cancel the common factor (w + 5).

$$=-\frac{2(w-5)}{3(w-2)}$

Now substitute this in the given expression.

$$\frac{50-2 w^{2}}{3 w^{2}+9 w-30} \cdot \frac{w^{2}+5 w-14}{6 w-30}=-\frac{2(w-5)}{3(w-2)} \cdot \frac{w^{2}+5 w-14}{6 w-30}$

Multiply the fractions $\frac{a}{b} \cdot \frac{c}{d}=\frac{a \cdot c}{b \cdot d}$

$$=-\frac{2(w-5)\left(w^{2}+5 w-14\right)}{3(w-2)(6 w-30)}$

Factor the denominator $3(w-2)(6 w-30) =18(w-2)(w-5)$

$$=-\frac{2(w-5)\left(w^{2}+5 w-14\right)}{18(w-2)(w-5)}$

Cancel the common factor 2(w – 5).

$$=-\frac{w^{2}+5 w-14}{9(w-2)}$

Factor the numerator $w^{2}+5 w-14=(w-2)(w+7)$

$$=-\frac{(w-2)(w+7)}{9(w-2)}$

Cancel the common factor (w – 2).

$$=-\frac{w+7}{9}$

$$\frac{50-2 w^{2}}{3 w^{2}+9 w-30} \cdot \frac{w^{2}+5 w-14}{6 w-30}=-\frac{w+7}{9}$

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