<span>We have 2 given numbers:
=> 86 + 68 + 38 = 192
=> 68 + 38 + 86 = 192 also
This kind of process used commutative property system of equation in where:
a + b = b + a
No matter what sequence you are going to use, the answer would always be the
same.
In the given 2 equations, notice that both digits of 2 group of equation
contains the same number. The only difference is they were added with different
orders. Yet the answers are still the same.
</span>
Step-by-step explanation:
<h3><u>Given</u><u>:</u><u>-</u></h3>
(√3+√2)/(√3-√2)
<h3><u>To </u><u>find</u><u>:</u><u>-</u></h3>
<u>Rationalised</u><u> form</u><u> </u><u>=</u><u> </u><u>?</u>
<h3><u>Solution</u><u>:</u><u>-</u></h3>
We have,
(√3+√2)/(√3-√2)
The denominator = √3-√2
The Rationalising factor of √3-√2 is √3+√2
On Rationalising the denominator then
=>[(√3+√2)/(√3-√2)]×[(√3+√2)/(√3+√2)]
=>[(√3+√2)(√3+√2)]×[(√3-√2)(√3+√2)]
=>(√3+√2)²/[(√3-√2)(√3+√2)]
=> (√3+√2)²/[(√3)²-(√2)²]
Since (a+b)(a-b) = a²-b²
Where , a = √3 and b = √2
=> (√3+√2)²/(3-2)
=> (√3-√2)²/1
=> (√3+√2)²
=> (√3)²+2(√3)(√2)+(√2)²
Since , (a+b)² = a²+2ab+b²
Where , a = √3 and b = √2
=> 3+2√6+2
=> 5+2√6
<h3><u>Answer:-</u></h3>
The rationalised form of (√3+√2)/(√3-√2) is 3+2√6+2.
<h3>
<u>Used formulae:-</u></h3>
→ (a+b)² = a²+2ab+b²
→ (a-b)² = a²-2ab+b²
→ (a+b)(a-b) = a²-b²
→ The Rationalising factor of √a-√b is √a+√b
Answer:
i need help please with my maths
Step-by-step explanation:
Answer:
<em>$11.21</em>
Step-by-step explanation:
$30.00 - $18.79 = $11.21
