Advanced Placement (AP), 03.09.2020 01:01 gui00g7888888888888

Which of the following are benefits of well-named variables? Select two answers.

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Answer from: eguzmandpandoracom

Options:

(a) The program will be easier to read.

(b) The program will run faster.

(d) The program will use less resources.

(a) The program will be easier to read.

Explanation:

Naming variables properly is an essential thing to do when writing programs or even developing software applications. This involves making use of descriptive and logical names when declaring variables.

For example, to store the age (say 67) of a user in a variable, the following ways are possible:

i. $x = 67

ii. $name = 67

iii. $age = 67

From (i), simply using $x = 67 does not exactly tell that 67 is the age of some user. That makes it a bit difficult to understand the code.

Also, from (ii), using $name = 67 is quite misleading. The variable name $name already depicts that the variable should store the name of some sort. However, a number is supplied. This is not a good way of naming variables.

The third option, from (iii), however clearly depicts the intention of the line of code without having to think too much. This is a well-named variable.

In general, well-named variables;

(i) makes it pretty easy to read and thus understand the lines of code in a program

(ii) following from (i), makes it easier for other programmers to be able to debug such program.

Well-named variables however have nothing to do with the speed of the execution of the program neither do they have anything to do with the amount of resources that will be used by the program.

Answer from: Quest

You need to figure out at which x-value does the graph has a tangent line slope of -1.remember, the derivative is the slope of the tangent line. so find the derivative function and set it to equal -1 to find the x-value that will have a tangent line slope of 0.differentiate with the chain rule. $\begin{aligned} f(x) & = (0.5)^{2x} \\ f'(x) & = 0.5^{2x} \cdot \ln(0.5) \cdot (2x)' \\ f'(x) & = 0.5^{2x} \cdot \ln(0.5) \cdot 2 \\ f'(x) & = 2\ln(0.5) \cdot 0.5^{2x} \end{aligned}$ set the derivative to equal -1 and solve for x. $\begin{aligned} -1 & = 2\ln(0.5) \cdot 0.5^{2x} \\ -\frac{1}{2} & = \ln(0.5) \cdot 0.5^{2x} \\ -\frac{1}{2\ln(0.5)} & = 0.5^{2x} \end{aligned}$ take log base 10 of both sides. also recall that $\log_c(m^n) = n \log_c m$ . $\begin{aligned} \log\left[ -\frac{1}{2\ln(0.5)}\right] & = \log\left[0.5^{2x}\right] \\ \log\left[ -\frac{1}{2\ln(0.5)}\right] & = 2x \log(0.5) \\ \dfrac{\log\left[ -\frac{1}{2\ln(0.5)}\right]}{2\log(0.5)} & = x \\ x & \approx 0.2356 \end{aligned}$ so the x-coord of the point of tangency is $x \approx 0.2356$ .the y-coordinate of this point of tangency can be calculated using the original function: $\begin{aligned} f(x) & = (0.5)^{2x} \\ f(0.2356) & = (0.5)^{2 \cdot 0.2356} \\ & \approx 0.7213 \end{aligned}$ using point-slope form, the equation of the tangent line is $y - 0.7213 = -1(x-0.2356)$ the x-intercept of this tangent line is when the graph is at y=0. $\begin{aligned} 0 - 0.7213 & = -1(x-0.2356)\\ - 0.7213 & = -(x-0.2356) \\ 0.7213 & = x-0.2356\\ x & = 0.9569 \end{aligned}$ this rounds off to $x & = 0.957$ .choice c.

Answer from: Quest

answer; vehicle vin#mileage, a detailed description of work requested, vehicle owner contact phone number;

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Which of the following are benefits of well-named variables? Select two answers....