Answer:
Step-by-step explanation:
Given: quadrilateral ABCD inscribed in a circle
To Prove:
1. ∠A and ∠C are supplementary.
2. ∠B and ∠D are supplementary.
Construction : Join AC and BD.
Proof: As, angle in same segment of circle are equal.Considering AB, BC, CD and DA as Segments, which are inside the circle,
∠1=∠2-----(1)
∠3=∠4-----(2)
∠5=∠6-------(3)
∠7=∠8------(4)
Also, sum of angles of quadrilateral is 360°.
⇒∠A+∠B+∠C+∠D=360°
→→∠1+∠2+∠3+∠4+∠5+∠6+∠7+∠8=360°→→→using 1,2,3,and 4
→→→2∠1+2∠4+2∠6+2∠8=360°
→→→→2( ∠1 +∠6) +2(∠4+∠8)=360°⇒Dividing both sides by 2,
→→→∠B + ∠D=180°as, ∠1 +∠6=∠B , ∠4+∠8=∠B------(A)
As, ∠A+∠B+∠C+∠D=360°
∠A+∠C+180°=360°
∠A+∠C=360°-180°------Using A
∠A+∠C=180°
Hence proved.
credit: someone else
29293
Step-by-step explanation:
yes yes !!
The mean is 52 because 42+45+58+63=208 then you divide it by the amount of numbers: 4 : 208 divided by 4= 52
Answer:
Step-by-step explanation:
Okay, so I think I know what the equations are, but I might have misinterpreted them because of the syntax- I think when you ask a question you can use the symbols tool to input it in a more clear way, otherwise you can use parentheses and such.
Problem 1:
(x²)/4 +y²= 1
y= x+1
*substitute for y*
Now we have a one-variable equation we can solve-
x²/4 + (x+1)² = 1
x²/4 + (x+1)(x+1)= 1
x²/4 + x²+2x+1= 1
*subtract 1 from both sides to set equal to 0*
x²/4 +x^2+2x=0
x²/4 can also be 1/4 * x²
1/4 * x² +1*x² +2x = 0
*combine like terms*
5/4 * x^2+2x+ 0 =0
now, you can use the quadratic equation to solve for x
a= 5/4
b= 2
c=0
the syntax on this will be rough, but I'll do my best...
x= (-b ± √(b²-4ac))/(2a)
x= (-2 ±√(2²-4*(5/4)*(0))/(2*(5/4))
x= (-2 ±√(4-0))/(2.5)
x= (-2±2)/2.5
x will have 2 answers because of ±
x= 0 or x= 1.6
now plug that back into one of the equations and solve.
y= 0+1 = 1
y= 1.6+1= 2.6
Hopefully this explanation was enough to help you solve problem 2.
Problem 2:
x² + y² -16y +39= 0
y²- x² -9= 0
