If two resistors with resistances R1 and R2 are connected in parallel, as in the figure below, then the total resistance R, measured in ohms (Ω), is given by 1/R = 1/R1 + 1/R2 . If R1 and R2 are increasing at rates of 0.3 Ω/s and 0.2 Ω/s, respectively, how fast is R changing when R1 = 60 Ω and R2 = 80 Ω? (Round your answer to three decimal places.)

The rate of change of R with time in the given equation is 0.004 ohm/s

Given parameters:

To find:

First determine the value of R from the given equation;

Finally, to determine the rate of change of R, differentiate the given equation.

Thus, from the given equation the rate of change of R with time is 0.004 ohm/s

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b. an equilateral triangle inscribed in a circle

step-by-step explanation:

we are given the figure of two circles having a triangle inside.

now, we know that,

"when a figure is lying inside an another figure, then the outside figure is known as circumscribed figure and the figure lying inside is referred to as the inscribed figure".

as we see that, the circles are containing the triangle completely.

thus, the circles are the circumscribed figures and the triangle is the inscribed figure.

hence, we get, 'an equilateral triangle is inscribed in a circle'.