Answer:
total = 0;
for (k = 0; k <= n; k++)
total += Math.pow(k,3);
Step-by-step explanation:
Here the variable total is declared and initialized with a value zero
Then a for loop is defined with a counter k whose initial value is set to zero then a condition for the loop is that the counter k does not exceed n k<=n and then within the loop a statement which add the cube of the counter k to the variable total and still assigns it to the variable total is defined
total += Math.pow(k,3);
What this program does is to obtain the sum of the cubes of k
Step-by-step explanation:
÷6
First, simplify the 6 to an improper fraction, which is:
Next rewrite the equation using the improper fraction:
÷
Now, in order to divide the fraction, we will need to flip the second fraction over, and then multiply the two fractions together. A great way to remember this is the phrase <em>Copy Dot Flip - </em>Copy the first fraction, put a dot for multiplication, and flip the second fraction.
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Finally, multiply together and simplify/reduce if needed:
<u>The final answer is </u>
<u> and no simplification or reduction is needed</u>
999 - 100 + 1 = 900
There are 900 numbers between 100 and 999 (inclusive)
We group the numbers into groups of 3s. So we find one number divisible by 3 in every group.
Number of numbers that can be divided by 3 = 900 ÷ 3 = 300
Number of numbers that can be divided by 3 = 300
We group the numbers into groups of 4s. So we find one number divisible by 4 in every group.
Number of numbers that can divided by 4 = 900 ÷ 4 = 225
Number of numbers that can divided by 4 = 225
Among the group of 4s, there is 1/3 that are already inclusive in divisible by 3.
225 ÷ 3 = 75
Number of numbers that can divided by 4 = 225 - 75 = 150
Total numbers of positive integers divisible by three and fours = 300 + 150 = 450
