Notice that
11/12 = 1/6 + 3/4
so that
tan(11π/12) = tan(π/6 + 3π/4)
Then recalling that
sin(x + y) = sin(x) cos(y) + cos(x) sin(y)
cos(x + y) = cos(x) cos(y) - sin(x) sin(y)
⇒ tan(x + y) = (tan(x) + tan(y))/(1 - tan(x) tan(y))
it follows that
tan(11π/12) = (tan(π/6) + tan(3π/4))/(1 - tan(π/6) tan(3π/4))
tan(11π/12) = (1/√3 - 1)/(1 + 1/√3)
tan(11π/12) = (1 - √3)/(√3 + 1)
tan(11π/12) = - (√3 - 1)²/((√3 + 1) (√3 - 1))
tan(11π/12) = - (4 - 2√3)/2
tan(11π/12) = - (2 - √3) … … … [A]
If we are eliminating x, we would multiply the 2nd equation by -5
5x + 18y = 23
x + 3y = 2....multiply by -5
-----------------
5x + 18y = 23
-5x - 15y = - 10 (result of multiplying by -5)
----------------add
0 + 3y = 13.....bye-bye x :)
Answer:
6 feet
Step-by-step explanation:
Since there are 8 bricks and each brick is 9 inches long, that is a total of 72 inches. Now, you need to convert it from inches to feet. Since 1 foot is 12 inches, we need to divide 72 by 12 to get the number of feet. 72 / 12 is equal to 6.
Answer
school building, so the fourth side does not need Fencing. As shown below, one of the sides has length J.‘ (in meters}. Side along school building E (a) Find a function that gives the area A (I) of the playground {in square meters) in
terms or'x. 2 24(15): 320; - 2.x (b) What side length I gives the maximum area that the playground can have? Side length x : [1] meters (c) What is the maximum area that the playground can have? Maximum area: I: square meters
Step-by-step explanation:
Answer:
Uuh whats the problem
Step-by-step explanation:
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