The 34th term of the sequence 158, 155, 152 ...... is 59
The sequence is as follows:
158, 155, 152 ......
let
a = first term = 158
d = common difference = 155 - 158 = -3
Using Arithmetic progression formula, let's find the 34th term as follows:
aₙ = a + (n - 1)d
a₃₄ = 158 + (34 - 1)-3
a₃₄ = 158 + (33)(-3)
a₃₄ = 158 - 99
a₃₄ = 59
Therefore, the 34th term of the sequence is 59.
learn more about sequence here: brainly.com/question/23251262?referrer=searchResults
You must graph the two given lines and find their point of intersection. It is (10,5). Shade the areas BELOW each of the 2 lines. One line intersects the y-axis at (0,15) and the other line intersects the x-axis at (12.5,0)
Thus, you have an area defined by the vertices given above.
What to do next? Steal the coordinates of each point and subst. them into the given objective function P = 15x + 20y. For exampel, for (12.5,0), the value of P is 15(12.5) + 20(0) = 187.5.
Find P for each of the remaining 3 vertices.
The largest value of P is the answer to this question.
Answer:
sine(11π/12): 0.2588190451
Step-by-step explanation:
You just basically put it into your calculator.
Make sure that is switched into radian mod.
0.2588190451
The answers is 25
5^2 is equal to 5*5=25
