The division equation that has a quotient of 13 is 1/1/13 = 13
<h3>How to determine the equation?</h3>
The given parameters are:
- Quotient = 13
- Dividend = Whole number
- Divisor = Unit fraction i.e. 1/n where n is an integer.
A division equation is represented as:
Dividend/Divisor = Quotient
Substitute 13 for the Quotient
Dividend/Divisor = 13
Recall that:
Unit fraction = 1/n
So, we have:
Dividend/1/n = 13
Let n = 13.
So, we have:
Dividend/1/13 = 13
This gives
13 * Dividend = 13
Divide both sides by 13
Dividend = 1
So, we have:
1/1/13 = 13
Hence, the division equation is 1/1/13 = 13
Read more about division equations at:
brainly.com/question/1622425
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Answer:
Quadrant 1 (1,5)
Step-by-step explanation:
Answer:
The answer is "MS and QS".
Step-by-step explanation:
Given ΔMNQ is isosceles with base MQ, and NR and MQ bisect each other at S. we have to prove that ΔMNS ≅ ΔQNS.
As NR and MQ bisect each other at S
⇒ segments MS and SQ are therefore congruent by the definition of bisector i.e MS=SQ
In ΔMNS and ΔQNS
MN=QN (∵ MNQ is isosceles triangle)
∠NMS=∠NQS (∵ MNQ is isosceles triangle)
MS=SQ (Given)
By SAS rule, ΔMNS ≅ ΔQNS.
Hence, segments MS and SQ are therefore congruent by the definition of bisector.
The correct option is MS and QS
At least 3. Have a nice day! :)
