Supplementary angles, when added, equal 180 degrees
so < A + < B = 180...and if < A = 80
80 + < B = 180
< B = 180 - 80
< B = 100 <==
(3 cos x-4 sin x)+(3sin x+4 cos x)=5
(3cos x+4cos x)+(-4sin x+3 sin x)=5
7 cos x-sin x=5
7cos x=5+sin x
(7 cos x)²=(5+sinx)²
49 cos²x=25+10 sinx+sin²x
49(1-sin²x)=25+10 sinx+sin²x
49-49sin²x=25+10sinx+sin²x
50 sin² x+10sinx-24=0
Sin x=[-10⁺₋√(100+4800)]/100=(-10⁺₋70)/100
We have two possible solutions:
sinx =(-10-70)/100=-0.8
x=sin⁻¹ (-0.8)=-53.13º (360º-53.13º=306.87)
sinx=(-10+70)/100=0.6
x=sin⁻¹ 0.6=36.87º
The solutions when 0≤x≤360º are: 36.87º and 306.87º.
Answer:
Is this the full question?
Step-by-step explanation:
You can find counterexamples to disprove this claim. We have positive integers that are perfect square numbers; when we take the square root of those numbers, we get an integer.
For example, the square root of 1 is 1, which is an integer. So if y = 1, then the denominator becomes an integer and thus we get a quotient of two integers (since x is also defined to be an integer), the definition of a rational number.
Example: x = 2, y = 1 ends up with 
which is rational. This goes against the claim that 
is always irrational for positive integers x and y.
Any integer y that is a perfect square will work to disprove this claim, e.g. y = 1, y = 4, y= 9, y = 16. So it is not always irrational.
Answer:
The equation of the circle is (x - 2)² + (y + 5)² = 144 ⇒ A
Step-by-step explanation:
The form of the equation of the circle is (x - h)² + (y - k)² = r², where
- r is the radius of the circle
- h, k are the coordinates of the center of the circle
Let us solve the question
∵ The center of the circle is at (2, -5)
→ From the rule above
∴ h = 2 and k = -5
∵ The radius of the circle is 12
∴ r = 12
→ Substitute the values of r, h, and k in the form of the equation above
∵ (x - 2)² + (y - -5)² = (12)²
∴ (x - 2)² + (y + 5)² = 144
∴ The equation of the circle is (x - 2)² + (y + 5)² = 144
