124° Is your answer.
The triangle Is an isosceles triangle which means the two bases will be the same angle. 180° is this sum of all angles in a triangle. If 62 is the right base angle then 62 must be the left base angle since is an isosceles triangle. Add 62 and 62 together then subtract from 180. That would give you 54° as the third angle angle at top. Since you are looking for a Z, we now know that z and 54° are symmetry and equal 180 when added together. We then subtract 54 from 180 to get the angle of z.
Answer:
The sequence is a function
Step-by-step explanation:
Given
Sequence: 
; 
Required
Is the sequence a function
To determine if the sequence is a function or not, we start by calculating its range from the domain:
When n = 0;
When n = 1: 
When n = 2; 
When n = 3; 
This implies that:
The domain (n) and the range (An) can be represented as:
![\left[\begin{array}{c}n&-&0&1&2&3\\-&-&-&\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dn%26-%260%261%262%263%5C%5C-%26-%26-%26%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{c}A_n&-&100&175&250&325\\-&-&-&\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7DA_n%26-%26100%26175%26250%26325%5C%5C-%26-%26-%26%5Cend%7Barray%7D%5Cright%5D)
<em>The above shows that the sequence is a one to one relation and a one to one relation is always a function.</em>
A squared plus b squared equals c squared. 5 squared is 25. 2 squared is 4. 25 plus 4 is 29. The square root of 29 is 5.385. Therefore, x=5.385. The equation for this would be •square root symbol• (25+4).
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:
m = - 
, c = - 
Step-by-step explanation:
The equation of a line in slope- intercept form is
y = mx + c ( m is the slope and c the y- intercept )
Given
2x + 3y + 1 = 0 ( subtract 2x + 1 from both sides )
3y = - 2x - 1 ( divide terms by 3 )
y = - x - ← in slope- intercept form
with m = - and c = -
