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/cm²where 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) $75 \int_{0}^{20} \frac{\sqrt{100-D^{2}}}{D+4} d D$

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The Answer of 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/cm²where 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 Thin Strip of Nutrients 20 Cm Long Is Placed 75D+4\frac{75}{D+4}D+475​

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