The number of bees spotted in Amelie's garden can also be modeled by the function √ B(x) = 50 k + 2x where x is the daily high temperature, in degrees Farenheit, and k is a positive constant. When the number of bees spotted is 100, the daily high temperature is increasing at a rate of 2◦F per day. According to this model, how quickly is the number of bees changing with respect to time when 100 bees are spotted?

According to this model, the rate of increase of bees spotted with respect to time when 100 bees are spotted = 80 bees per day.

Step-by-step explanation:

√B(x) = 50 k + 2x

where x is the daily high temperature, in degrees Farenheit.

k is a positive constant.

When the number of bees spotted = 100,

The daily high temperature is increasing at a rate of 2°F per day; (dx/dt) = 2 °F/day

How quickly is the number of bees changing with respect to time when 100 bees are spotted?

√B(x) = 50k + 2x

Taking a time derivative of both sides,

(d/dt) [√B] = (d/dt) [50 k + 2x]

(1/2) (B⁻⁰•⁵) (dB/dt) = (d/dt)(50k) + (d/dt)(2x)

½(B⁻⁰•⁵) (dB/dt) = 50 (dk/dt) + 2 (dx/dt)

B = 100 bees

(dB/dt) = ?

(dk/dt) = 0 (since k is a positive constant)

(dx/dt) = 2°F/day

½(100⁻⁰•⁵) (dB/dt) = 50 (0) + 2 (2)

½(10⁻¹) (dB/dt) = 4

(1/2)(1/10) (dB/dt) = 4

(dB/dt) = 4×2×10 = 80

(dB/dt) = 80 bees per day

Hope this Helps!!!

3/4 ÷ 3/4

step-by-step explanation:

basically divide any proper fraction with itself to fit the criteria.

so if we are using 3/4, then:

3/4 ÷ 3/4 = 1

so 1 is the quotient.

and the numerators for both these fractions is 3.   and if we divide the numerators:

3 ÷ 3 = 1

we see that quotient is equal to our original quotient as well.

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