108 is the answer for this question
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Answer:
nπ -π/6 . . . for any integer n
Step-by-step explanation:
tan(x) +√3 = -2tan(x) . . . . . given
3tan(x) = -√3 . . . . . . . . . . . add 2tan(x)-√3
tan(x) = -√3/3 . . . . . . . . . . divide by 3
x = arctan(-√3/3) = -π/6 . . . . use the inverse tangent function to find x
This is the value in the range (-π/2, π/2). The tangent function repeats with period π, so the set of values of x that will satisfy this equation is ...
x = n·π -π/6 . . . . for any integer n
It’s too small zoom in some I will be glad to assist you
Answer:
(x1,y1) = (2x - x2, 2y - y2)
Step-by-step explanation:
Given:
Midpoint = (x , y)
End point = (x2, y2)
Find:
(x1, y1)
Computation:
Mid-point formula
x = (x1 + x2) / 2 , y = (y1 + y2) / 2
So,
2x = x1 + x2 , 2y = y1 + y2
x1 = 2x - x2 , y1 = 2y - y2
So,
(x1,y1) = (2x - x2, 2y - y2)
Okay Let's say
2x + 3 = 45
-3 -3
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2x = 42
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2 2 x = 21