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Verdich [7]
3 years ago
14

If y varies directly with x and y = 3 when x =12, then what is the value of y when x =40?

Mathematics
2 answers:
3241004551 [841]3 years ago
4 0
Its c 10 and other fillar so i can answer
BlackZzzverrR [31]3 years ago
3 0
The answer is c (10) because the value of y is 10
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Assume that x and y are both differentiable functions of t and find the required values of dy/dt and dx/dt.
Sedbober [7]

Answer:

a. \frac{dy}{dt} = -\frac{13}{8}

b. \frac{dx}{dt} = \frac{9}{2}

Step-by-step explanation:

To solve this question, we apply implicit differentiation.

xy = 2

Applying the implicit differentiation:

y\frac{dx}{dt} + x\frac{dy}{dt} = \frac{d}{dt}(2)

y\frac{dx}{dt} + x\frac{dy}{dt} = 0

a. Find dy/dt, when x = 4, given that dx/dt = 13.

x = 4

So

xy = 2

4y = 2

y = \frac{2}{4} = \frac{1}{2}

Then

y\frac{dx}{dt} + x\frac{dy}{dt} = 0

\frac{1}{2}(13) + 4\frac{dy}{dt} = 0

4\frac{dy}{dt} = -\frac{13}{2}

\frac{dy}{dt} = -\frac{13}{8}

b. Find dx/dt, when x = 1, given that dy/dt = -9.

x = 1

So

xy = 2

y = 2

Then

y\frac{dx}{dt} + x\frac{dy}{dt} = 0

2\frac{dx}{dt} - 9 = 0

2\frac{dx}{dt} = 9

\frac{dx}{dt} = \frac{9}{2}

3 0
3 years ago
Solve y=x+3 and 2x+y=-6 by substitution
allsm [11]
Hope this is right....

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3 years ago
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