Answer:
Defining relations for tangent, cotangent, secant, and cosecant in terms of sine and cosine
Step-by-step explanation:
<h2>HOPE THIS HELPS IM SINGLE 13 STRAIGHT AND IMMA GIRL ;)</h2>
Answer:
2550/6= 425
he has to save 425 each month
Step-by-step explanation:
Answer:
n =
, n = 
Step-by-step explanation:
6n² - 5n - 7 = - 8 ( add 8 to both sides )
6n² - 5n + 1 = 0 ← in standard form
Consider the product of the factors of the coefficient of the n² term and the constant term which sum to give the coefficient of the n- term
product = 6 × 1 = 6 and sum = - 5
The factors are - 3 and - 2
Use these factors to split the n- term
6n² - 3n - 2n + 1 = 0 ( factor the first/second and third/fourth terms )
3n(2n - 1) - 1(2n - 1) = 0 ← factor out (2n - 1) from each term
(2n - 1)(3n - 1) = 0 ← in factored form
Equate each factor to zero and solve for n
3n - 1 = 0 ⇒ 3n = 1 ⇒ n = 
2n - 1 = 0 ⇒ 2n = 1 ⇒ n = 
Answer:
69
Step-by-step explanation:
so x2 will give you 9 times 7 which is 63. add it to 2 times 3 which is 63 + 6 will give you 69.
Answer:
b
Step-by-step explanation:
In general
Given
y = f(x) then y = f(Cx) is a horizontal stretch/ compression in the x- direction
• If C > 1 then compression
• If 0 < C < 1 then stretch
Consider corresponding points on the 2 graphs
(2, 2 ) → (4, 2 )
(4, - 2 ) → (8, - 2 )
Indicating a stretch in the x- direction.
y = f(
) with C =
, that is 0 < C < 1
stretches the graph in the x- direction by a factor of 2
Thus
y = f(
) → b