2. so, e is a number like π. it is irrational, so its easiest to express it with a symbol. however, it is…

2. so, e is a number like π. it is irrational, so its easiest to express it with a symbol. however, it is just a number, not a variable. having said that we can graph exponential functions that use e as a base. since e is between 2 and 3, lets compare the graphs of ( f(x)=3^{x} ), ( g(x)=e^{x} ), and ( h(x)=2^{x} ). so, which is which? what would the graph of ( y = e^{-x} ) look like? sketch it above. 3. answer the following, based off the graphs above. as ( x \to infty ), ( e^{x} \to )? as ( x \to -infty ), ( e^{x} \to )? as ( x \to infty ), ( e^{-x} \to )? as ( x \to -infty ), ( e^{-x} \to )?

2. so, e is a number like π. it is irrational, so its easiest to express it with a symbol. however, it is just a number, not a variable. having said that we can graph exponential functions that use e as a base. since e is between 2 and 3, lets compare the graphs of ( f(x)=3^{x} ), ( g(x)=e^{x} ), and ( h(x)=2^{x} ). so, which is which? what would the graph of ( y = e^{-x} ) look like? sketch it above. 3. answer the following, based off the graphs above. as ( x \to infty ), ( e^{x} \to )? as ( x \to -infty ), ( e^{x} \to )? as ( x \to infty ), ( e^{-x} \to )? as ( x \to -infty ), ( e^{-x} \to )?

Answer

Explanation:

Step1: Recall the formula for initial value

For an exponential function of the form (D(t)=a\cdot e^{kt}), the initial value (when (t = 0)) is found by substituting (t=0) into the function.

Step2: Substitute (t = 0) into (D(t)=35e^{-0.2t})

When (t = 0), we have (D(0)=35e^{-0.2\times0}). Since (e^{0}=1) (by the property of exponents (a^{0}=1) for (a\neq0), and (e\approx2.718\neq0)), then (D(0)=35\times1).

Answer:

35 milligrams