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Chapter 1: Problem 35

The variable a starts with the value 1 The variable b starts with the value 10 . The variable c starts with the value 100 . The variable \(x\) starts with the value 0 Store the value of \(c\) times 3 in \(x\). Add the value of b times 6 to the value already in \(x\). Add the value of a times 5 to the value already in \(x\). Display the value in \(x\) on the screen.

Short Answer

Expert verified
Answer: 365

Step by step solution

01

Initialize variables

We are given the following initial values for our variables: \(a = 1\), \(b = 10\), \(c = 100\), and \(x = 0\).
02

Multiply the value of c by 3

We need to calculate the product of \(c\) and \(3\). The expression for this is \(3c\). We can plug in the value of \(c\) to get: \(3 \times 100 = 300\).
03

Assign the value obtained in Step 2 to x

Now that we have the value of \(3c\), we can set x equal to this product, so we have: \(x = 300\).
04

Multiply the value of b by 6

Now, we need to calculate the product of \(b\) and \(6\). The expression for this is \(6b\). Plugging in the value of \(b\), we get: \(6 \times 10 = 60\).
05

Add the value obtained in Step 4 to the current value of x

We now need to add the value \(6b\) from Step 4 to the current value of \(x\) (that is \(300\) from Step 3): \(x = 300 + 60 = 360\).
06

Multiply the value of a by 5

Next, we will calculate the product of \(a\) and \(5\). The expression for this is \(5a\). We can plug in the value of \(a\) to get: \(5 \times 1 = 5\).
07

Add the value obtained in Step 6 to the current value of x

We now need to add the value \(5a\) from Step 6 to the current value of \(x\) (that is \(360\) from Step 5): \(x = 360 + 5 = 365\).
08

Display the value of x

The final value of \(x\) after all operations is \(365\). Display this value on the screen: \(x = 365\).

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Most popular questions from this chapter

Both main memory and secondary storage are types of memory. Describe the difference between the two.

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