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?

The value of at the instant when particle moves on the hyperbola from time is .

Differentiating the given equation   with respect to the to get an addition equation by  implicit differentiation,

Also, from the given equation, we can write

Substitute the value of in the equation (i), we get

Substitute the value of and in the above equation, we get

Now, from equation (i), we can re-write the expression to evaluate the value of at that instant as-

Hence, the value of at the instant when particle moves on the hyperbola from time is .

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The value of x at this instant is 3.

Step-by-step explanation:

Let , we get an additional equation by implicit differentiation:

(1)

From the first equation we find that:

(2)

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

(3)

If we know that and , then the first derivative of x in time is:

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

The value of x at this instant is 3.

the final price would be 29.6925

step-by-step explanation:

39.59 minus 25% equals 29.6925

x = -3

y = -18

step-by-step explanation:

to solve by substitution, substitute one variable and solve for the other. then substitute again to find the remaining variable.

x-1/3(3x-9) = -3 since y=3x-9 substittute y.

x-1/3(3x-9)=-3

x + x +3 = -3

2x+3=-3

2x+3-3= -3-3

2x=-6

x=-3

substitute again into y=3x-9

y=3(-3)-9

y=-9-9

y=-18