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Nina [5.8K]
3 years ago
11

perform the indicated operations a. 3/10+6/10. b. 1/3+2/4+1/6. c. 5/6-3/6 d. 2/3-6/10 e. 4/10×3/7 f. 1/6x6/15 g. 1/8÷4/9 h. 1/5÷

3/4​
Mathematics
1 answer:
babymother [125]3 years ago
7 0

Answer:

A. \frac{9}{10}

B. 1

C. \frac{2}{6} or \frac{1}{3}

D. \frac{2}{30} or \frac{1}{15}

E. \frac{12}{70} or \frac{6}{35}

F. \frac{6}{90} or \frac{1}{15}

G. \frac{9}{32}

H. \frac{4}{15}

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A particle moves on the hyperbola xy=18 for time t≥0 seconds. At a certain instant, y=6 and dydt=8. What is x that this instant?
professor190 [17]

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The value of x at this instant is 3.

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x\cdot \frac{dy}{dt}+y\cdot \frac{dx}{dt} = 0 (1)

From the first equation we find that:

x = \frac{18}{y} (2)

By applying (2) in (1), we get the resulting expression:

\frac{18}{y}\cdot \frac{dy}{dt}+y\cdot \frac{dx}{dt} = 0 (3)

y\cdot \frac{dx}{dt}=-\frac{18}{y}\cdot \frac{dy}{dt}

\frac{dx}{dt} = -\frac{18}{y^{2}} \cdot \frac{dy}{dt}

If we know that y = 6 and \frac{dy}{dt} = 8, then the first derivative of x in time is:

\frac{dx}{dt} = -\frac{18}{6^{2}} \cdot (8)

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From (1) we determine the value of x at this instant:

x\cdot \frac{dy}{dt} = -y\cdot \frac{dx}{dt}

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x = -6\cdot \left(\frac{-4}{8} \right)

x = 3

The value of x at this instant is 3.

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3 years ago
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