<h3>
Answer: y = 12/(x^2)</h3>
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Explanation:
"y varies inversely as x^2" means y = k/(x^2) for some constant k.
Plug in (x,y) = (2,3) and solve for k
y = k/(x^2)
3 = k/(2^2)
3 = k/4
3*4 = k
12 = k
k = 12
The original equation updates to y = 12/(x^2)
As a check, plugging in x = 2 should lead to y = 3
y = 12/(x^2)
y = 12/(2^2)
y = 12/4
y = 3 ... we get the proper y value, so the answer is confirmed.
Answer:
918
Step-by-step explanation:
Given data
Initila population= 85
Rate = 12%
TIme= 21 years
Let us apply the compounding expression
A=P(1+r)^t
substitute
A= 85(1+0.12)^21
A=85(1.12)^21
A=85*10.80
A= 918
Hence the approximate Rabbit population will be 918
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!!
Answer:
-1=7-8
Step-by-step explanation:
-1=7-8
-1=-1
<em>Look</em><em> </em><em>at</em><em> </em><em>the</em><em> </em><em>attached</em><em> </em><em>picture</em>
<em>H</em><em>ope</em><em> </em><em>it</em><em> </em><em>will</em><em> </em><em>be</em><em> </em><em>helpful</em><em> </em><em>to</em><em> </em><em>you</em><em>.</em><em>.</em><em>.</em><em>.</em>
<h3>
<em>Good</em><em> </em><em> </em><em>luck</em><em> </em><em>on </em><em>your</em><em> </em><em>as</em><em>signment</em></h3>
<em>-pragya~</em><em>~</em>