Answer:
Equation: 3x + 2 = 50
# of friends: 16
Step-by-step explanation:
3 chocolates each or 3x because what "each" means is not known yet-
3x
<em>Note: Always use x for an unknown value</em>
2 remaining chocolates-
3x + 2
50 total chocolates-
3x + 2 = 50
Remove two to keep the x alone-
3x = 48
Remove the 3 or the "Coefficient" by dividing-
48/3 = 16
x = 16
Equation: 3x + 2 = 50
# of friends: 16
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:
I=$ 198
Step-by-step explanation:
I=PRT
P=1200
T=3
R=5.5% = 0.055
I=1200*0.055*3= 198
Answer:
D) 2x<u>2</u>
Step-by-step explanation:
i got i right on the test
Write each line to make the sum 7 it say I can't tell y'all the answer dang alot of people need help I got 18 seconds to tell yall
