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A Thin Strip of Nutrients 20 Cm Long Is Placed 75D+4\frac{75}{D+4}

Question 5

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A thin strip of nutrients 20 cm long is placed in a circular petri dish of radius 10 cm, as shown.The population density of bacteria in the disk after 3 hours is given by 75D+4\frac{75}{D+4} bacteria/cm2 where D is the distance (in cm) to the nutrient strip.Which of the following integrals gives the number of bacteria in the petri dish 3 hours after the nutrient strip has been introduced?  A thin strip of nutrients 20 cm long is placed in a circular petri dish of radius 10 cm, as shown.The population density of bacteria in the disk after 3 hours is given by  \frac{75}{D+4}  bacteria/cm<sup>2 </sup>where D is the distance (in cm) to the nutrient strip.Which of the following integrals gives the number of bacteria in the petri dish 3 hours after the nutrient strip has been introduced?   A)   150 \pi \int_{0}^{20} \frac{100-D^{2}}{D+4} d D  B)   75 \pi \int_{0}^{3} \frac{\sqrt{100-D^{2}}}{D+4} d D  C)   300 \int_{0}^{10} \frac{\sqrt{100-D^{2}}}{D+4} d D  D)   75 \int_{0}^{20} \frac{\sqrt{100-D^{2}}}{D+4} d D


A) 150π020100D2D+4dD150 \pi \int_{0}^{20} \frac{100-D^{2}}{D+4} d D
B) 75π03100D2D+4dD75 \pi \int_{0}^{3} \frac{\sqrt{100-D^{2}}}{D+4} d D
C) 300010100D2D+4dD300 \int_{0}^{10} \frac{\sqrt{100-D^{2}}}{D+4} d D
D) 75020100D2D+4dD75 \int_{0}^{20} \frac{\sqrt{100-D^{2}}}{D+4} d D

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