Answer:
Hello,
Step-by-step explanation:
We divide the interval [a b] in n equal parts.



<span>A
student divided the given number by 100 that resulted to 28.003 in his chart.
So what we are looking for this situation is the value of the number before it
was divided by 100.
To get the value, simply multiply the result by 100
=> 28.003 x 100
=> 2800.3
by dividing 100 to the given number, we
simply move the value to the right in a value of hundreds.
See attached Image.
</span>
2(2x +2y) and <span>√(4x + 4y)^2</span>
It is probably easier to open an account with a bank rather than with a credit union because most credit unions require a kind of affiliation, but banks will let anyone with money to open an account. The correct option among all the options that are given in the question is the first option or option "a".
Hope this helps!