Answer:
The man rushes through your door with his fists in the air, all of the sudden your vision went black, and he had stolen your points.
Step-by-step explanation:
Use long division if you have to.
The answer is 60
The value of x + y is 12
<h3>How to solve for x + y?</h3>
The equation is given as:
x^3 - y^3 = 218
Make x^3 the subject
x^3 = 218 + y^3
The next cubic expression greater than 218 is 343.
So, we assume
x^3 = 343
This gives
x = 7
Substitute x^3 = 343 in x^3 = 218 + y^3
343 = 218 + y^3
Evaluate the like terms
y^3 = 125
This gives
y = 5
So, we have:
x + y = 7 + 5
Evaluate
x + y = 12
Hence, the value of x + y is 12
Read more about equations at:
brainly.com/question/2972832
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Step-by-step explanation:
<h3><u>Given :-</u></h3>
[1+(1/Tan²θ)] + [ 1+(1/Cot²θ)]
<h3>
<u>Required To Prove :-</u></h3>
[1+(1/Tan²θ)]+[1+(1/Cot²θ)] = 1/(Sin²θ-Sin⁴θ)
<h3><u>Proof :-</u></h3>
On taking LHS
[1+(1/Tan²θ)] + [ 1+(1/Cot²θ)]
We know that
Tan θ = 1/ Cot θ
and
Cot θ = 1/Tan θ
=> (1+Cot²θ)(1+Tan²θ)
=> (Cosec² θ) (Sec²θ)
Since Cosec²θ - Cot²θ = 1 and
Sec²θ - Tan²θ = 1
=> (1/Sin² θ)(1/Cos² θ)
Since , Cosec θ = 1/Sinθ
and Sec θ = 1/Cosθ
=> 1/(Sin²θ Cos²θ)
We know that Sin²θ+Cos²θ = 1
=> 1/[(Sin²θ)(1-Sin²θ)]
=> 1/(Sin²θ-Sin²θ Sin²θ)
=> 1/(Sin²θ - Sin⁴θ)
=> RHS
=> LHS = RHS
<u>Hence, Proved.</u>
<h3><u>Answer:-</u></h3>
[1+(1/Tan²θ)]+[1+(1/Cot²θ)] = 1/(Sin²θ-Sin⁴θ)
<h3><u>Used formulae:-</u></h3>
→ Tan θ = 1/ Cot θ
→ Cot θ = 1/Tan θ
→ Cosec θ = 1/Sinθ
→ Sec θ = 1/Cosθ
<h3><u>Used Identities :-</u></h3>
→ Cosec²θ - Cot²θ = 1
→ Sec²θ - Tan²θ = 1
→ Sin²θ+Cos²θ = 1
Hope this helps!!
