The x values range from -6 to 6, *including* those two numbers, so the domain is [-6, 6], and the y values range from -2 to 2, also inclusively, so the range is [-2, 2]
Answer:
The short answer is there isn’t.
Start by writing each of these as an expression:
x * y = 60
x + y = 7
Next, solve each for the same variable; in this case, y:
(x * y) / x = 60 / x
.: y = 60 / x
(x + y) - x = 7 - x
.: y = 7 - x
Next, replace y of the second expression to the first
y = 60 / x & y = 7 - x
.: 7 - x = 60 / x
Now, solve for x:
(7 - x) * x = (60 / x) * x
.: x * 7 - x^2 = 60
This is quadratic, so write it in the form of ax2 + bx + x = 0
(-1)x^2 + (7)x + (-60) = 0
.: a = -1, b = 7, c = -60
Finally solve for b:
x = (-b +- sqrt(b^2 - 4*a*c)) / 2a
.: x = (-7 +- sqrt(7^2 - 4*-1*-60)) / (2 * -1)
.: x = (-7 +- sqrt(49 - 240)) / -2
.: x = (-7 +- sqrt(-191)) / -2
The square root of a negative value is imaginary and thus there’s no real answer to this problem.
First, we are given that the inscribed angle of arc CB which is angle D is equal to 65°. This is half of the measure of the arc which is equal to the measure of the central angle, ∠O.
m∠O = 2 (65°) = 130°
Also, the measure of the angles where the tangent lines and the radii meet are equal to 90°. The sum of the measures of the angle of a quadrilateral ACOB is equal to 360°.
m∠O + m∠C + m∠B + m∠A = 360°
Substituting the known values,
130° + 90° + 90° + m∠A = 360°
The value of m∠A is equal to 50°.
<em>Answer: 50°</em>
