Question

two petri dishes were on the desktop in dr. mcfarlarians lab. dish a had cells with a radius of 4.6×10⁻¹³ cm. dish b had cells that had a radius of 3.9×10⁻¹¹ cm. what is the sum of the radii of the two types of cells, using scientific notation? a. 394.6×10⁻¹³ cm b. 3.946×10⁻¹⁵ cm c. 3.946×10⁻¹¹ cm

Explanation:

Step1: Convert to same exponent

To add numbers in scientific notation, we first make the exponents the same. For Dish B's radius \(3.9\times10^{-11}\) cm, we can rewrite it as \(390\times10^{-13}\) cm (since \(10^{-11}=100\times10^{-13}\), so \(3.9\times100 = 390\)). Dish A's radius is \(4.6\times10^{-13}\) cm.

Step2: Add the coefficients

Now add the coefficients: \(4.6 + 390=394.6\). So the sum is \(394.6\times10^{-13}\) cm. But we need to adjust to proper scientific notation. \(394.6\times10^{-13}=3.946\times10^{2}\times10^{-13}=3.946\times10^{-11}\) cm? Wait, no, wait. Wait, \(3.9\times10^{-11}=3.9\times10^{2}\times10^{-13}=390\times10^{-13}\). Then \(4.6\times10^{-13}+390\times10^{-13}=(4.6 + 390)\times10^{-13}=394.6\times10^{-13}\). Now, convert \(394.6\times10^{-13}\) to proper scientific notation: \(394.6 = 3.946\times10^{2}\), so \(3.946\times10^{2}\times10^{-13}=3.946\times10^{-11}\). Wait, let's check again. Wait, \(10^{-11}=10^{2}\times10^{-13}\), so \(3.9\times10^{-11}=3.9\times10^{2}\times10^{-13}=390\times10^{-13}\). Then \(4.6\times10^{-13}+390\times10^{-13}=394.6\times10^{-13}\). Now, \(394.6\times10^{-13}=3.946\times10^{2}\times10^{-13}=3.946\times10^{-11}\) (since \(2-13=-11\)). Wait, but option A is \(394.6\times10^{-13}\), option C is \(3.946\times10^{-11}\). Let's compute \(4.6\times10^{-13}+3.9\times10^{-11}\). Let's convert \(3.9\times10^{-11}\) to \(390\times10^{-13}\) (because \(10^{-11}=10^{2}\times10^{-13}\)), then add to \(4.6\times10^{-13}\): \(4.6 + 390 = 394.6\), so \(394.6\times10^{-13}\) cm. But \(394.6\times10^{-13}=3.946\times10^{2}\times10^{-13}=3.946\times10^{-11}\) cm. So the correct answer is C? Wait, let's calculate: \(4.6\times10^{-13}=0.046\times10^{-11}\). Then \(0.046\times10^{-11}+3.9\times10^{-11}=(0.046 + 3.9)\times10^{-11}=3.946\times10^{-11}\) cm. Ah, that's a better way. Convert Dish A's radius to \(10^{-11}\) exponent: \(4.6\times10^{-13}=4.6\times10^{-2}\times10^{-11}=0.046\times10^{-11}\). Then add to Dish B's \(3.9\times10^{-11}\): \(0.046 + 3.9 = 3.946\), so \(3.946\times10^{-11}\) cm. So option C is correct.

Answer:

C. \(3.946\times10^{-11}\) cm