If we take the square of x and square of y and then subtract them:
(csc t)²-(cot t)²=1 ( this eq. gets from basic identity
x²-y²=1......a 1+cot²x=csc²x)
equation 'a' represent the equation of hyperbola which is (x²/a²)-(y²/b²) =1 with given conditions( a=1,b=1)
So, option D is correct
B= 20°
and c = 160°
b is equal to the given angle, b = 20°
c is 180 - 20 = 160°
Answer:
length is 5
width is 4
Step-by-step explanation:
The solutions will lie in all the four quadrants.
y≤2/7x+1
We have the following inequality:
y=2/7x+1
If x=0
y=2/7(0) +1
y=1
If y=0
x=-3.5
So the line passes through the points:
(0,1) and (-3.5,0)
To find the shaded region, let us take a point, namely, the origin and test it in the inequality:
y≤ 2/7x +1
0≤ 2/7(0) +1
0≤ 1 which is true
Since this is true, then the shaded region includes this point. This is shown below and as you can see the solutions exist in all four quadrants.
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Answer:
mM = 113 and mN = 61
Step-by-step explanation:
In a Cyclic quadrilateral, the rule states that:
The sum opposite interior angles is equal to 180°
In the above diagram, we have cyclic quadrilateral KLMN
According to the rule stated above:
Angle K is Opposite to Angle M
So, Angle K + Angle M = 180°
Angle K = 67°
67° + Angle M = 180°
Angle M = 180° - 67°
Angle M = 113°
Angle L is Opposite to Angle N
so Angle L + Angle N = 180°
Angle L is given as = 119°
119° + Angle N = 180°
Angle N = 180° - 119°
Angle N = 61°
Therefore, Angle M = 113° and Angle N = 61°
