The sum of the two <em>rational</em> equations is equal to (3 · n² + 5 · n - 10) / (3 · n³ - 6 · n²).
<h3>How to simplify the addition between two rational equations</h3>
In this question we must use <em>algebra</em> definitions and theorems to simplify the addition of two <em>rational</em> equations into a <em>single rational</em> equation. Now we proceed to show the procedure of solution in detail:
- (n + 5) / (n² + 3 · n - 10) + 5 / (3 · n²) Given
- (n + 5) / [(n + 5) · (n - 2)] + 5 / (3 · n²) x² - (r₁ + r₂) · x + r₁ · r₂ = (x - r₁) · (x - r₂)
- 1 / (n - 2) + 5 / (3 · n²) Associative and modulative property / Existence of the multiplicative inverse
- [3 · n² + 5 · (n - 2)] / [3 · n² · (n - 2)] Addition of fractions with different denominator
- (3 · n² + 5 · n - 10) / (3 · n³ - 6 · n²) Distributive property / Power properties / Result
To learn more on rational equations: brainly.com/question/20850120
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Answer:
1:0.6
Step-by-step explanation:
1.b 2.34 is the correct answer
<h2><u>Part A:</u></h2>
Let's denote no of seats in first row with r1 , second row with r2.....and so on.
r1=5
Since next row will have 10 additional row each time when we move to next row,
So,
r2=5+10=15
r3=15+10=25
<u>Using the terms r1,r2 and r3 , we can find explicit formula</u>
r1=5=5+0=5+0×10=5+(1-1)×10
r2=15=5+10=5+(2-1)×10
r3=25=5+20=5+(3-1)×10
<u>So for nth row,</u>
rn=5+(n-1)×10
Since 5=r1 and 10=common difference (d)
rn=r1+(n-1)d
Since 'a' is a convention term for 1st term,
<h3>
<u>⇒</u><u>rn=a+(n-1)d</u></h3>
which is an explicit formula to find no of seats in any given row.
<h2><u>Part B:</u></h2>
Using above explicit formula, we can calculate no of seats in 7th row,
r7=5+(7-1)×10
r7=5+(7-1)×10 =5+6×10
r7=5+(7-1)×10 =5+6×10 =65
which is the no of seats in 7th row.
Answer:
1. (-5)
Step-by-step explanation:
Here you go, glad to help :)