TQRS is an inscribed quadrilateral.
5 x - 52° + 3 x + 40° = 180°
8 x - 12° = 180°
8 x = 180° + 12°
8 x = 192°
x = 192° : 8 = 24°
m∠ R = 3 · 24° + 40° = 112°
m∠ T = 5 · 24° - 52° = 68°
m∠ S = 360° - ( 68° + 68° + 112° ) = 112°
Answer:
m∠R, m∠S, m∠T = 112°, 112°, 68°.
9514 1404 393
Answer:
(x +6)^2 +(y -10)^2 = 225
Step-by-step explanation:
The standard form equation for a circle is ...
(x -h)^2 + (y -k)^2 = r^2
where the center is (h, k) and the radius is r.
The center of a circle is the midpoint of any diameter. The midpoint between two points is the average of their coordinates.
((-15, -2) +(3, 22))/2 = (-15+3, -2+22)/2 = (-6, 10)
The radius can be found using the distance formula, or by simply putting one of the given points in the equation for the circle to see what the constant (r^2) needs to be.
(x -(-6))^2 +(y -10)^2 = (-15-(-6))^2 +(-2-10)^2
(x +6)^2 +(y -10)^2 = 81 +144 = 225
The equation of the circle is ...
(x +6)^2 +(y -10)^2 = 225
The total area of pool and border = 361 square feet.
Length of square pool = x
Length of square pool plus border = (x + 1 +1 ) = x + 2
There is 1 foot length on each side on the x length.
The other length of the square pool and border = (x + 2)
Area = (x+2)(x+2)
(x + 2)(x +2) = 361. Let A = x +2
A*A = 361
A² = 361 Take square root of both sides
A = √361
A = 19
A = x + 2 = 19
x = 19 - 2
x = 17
Area of square pool = x*x = 17*17 = 289
Area of the pool = 289 square feet.
If the equation is 4x+1/3=20, your answer would be x=59/12.
- You would subtract 1/3 from each side. (1/3 - 1/3) (20-1/3)
- Then your new equation would be 4x = 59/3
- You would divide 4 from each side (4/4) (59/3 / 4)
- x = 59/12
