Answer:
is the required quotient.
Step-by-step explanation:
We have been given the expression:
Cancel the common term which is 3 we will be left with
can be written as after prime factorization is:
And 
can be written as after prime factorization is:
Hence, the given expression would become after 2 gets cancel from numerator and denominator:
Quotient is the answer we get after dividing numerator by denominator and cancel common term which is 
is the required quotient.
Answer:
A. Or 37/55
Step-by-step explanation:
Cross-cancel common factor=
37/5x1/11
Multiply fractions=
37x1/5x11
Multply the numbers=
37/55
<u><em>Feel free to mark brainliest!</em></u>
The program is an illustration of loops
<h3>What are loops?</h3>
Loops are program statements that are used to perform repetition
<h3>The main program</h3>
The program written in Python, where comments are used to explain each line is as follows:
#This initializes sum to 0
summ = 0
#This gets input for the first number
num = int(input())
#This is repeated while num is not -1
while num!= -1:
#This calculates the sum
summ+=num
#This gets input for the num
num = int(input())
#This prints the sum
print(summ)
Read more about loops at:
brainly.com/question/16397886
Answer:
Step-by-step explanation:
no ,x*x*x*x*x=x^5
x+x+x+x+x=5x
in multiplication those who have same base we add their power.
in addition those variables which have same base(or are like term) we add them.
Answer:
20
Step-by-step explanation:
It seems like the rule of this sequence is to add 5 (since 2 + 5 = 7 and 7 + 5 = 12). We already are given that we have 2 2-digit numbers (12 and 17) so let's see if there are any more. The sequence continues to 22, 27, 32, 37, 42, 47, ..., 92, 97. We need to count how many numbers are in the list 12, 17 ... 92, 97. To do this, let's add 3 to every term in the list to get 15, 20, ... 95, 100. Since the list is now full of multiples of 5 we can divide the list by 5 to get 3, 4, ... 19, 20 and then subtract 2 to get 1, 2, ... 17, 18 which means that there are 18 2-digit numbers.
