#1- I'm pretty sure that this one is the bottom right
#2- I'm pretty sure that this one is the top right
PLEASE CORRECT ME IF I'M WRONG!!!
1. Quadrilateral ABCD is inscribed in circle O
A quadrilateral is a four sided figure, in this case ABCD is a cyclic quadrilateral such that all its vertices touches the circumference of the circle.
A cyclic quadrilateral is a four sided figure with all its vertices touching the circumference of a circle.
2. mBCD = 2 (m∠A) = Inscribed Angle Theorem
An inscribed angle is an angle with its vertex on the circle, formed by two intersecting chords.
Such that Inscribed angle = 1/2 Intercepted Arc
In this case the inscribed angle is m∠A and the intercepted arc is MBCD
Therefore; m∠A = 1/2 mBCD
4. The sum of arcs that make up a circle is 360
Therefore; mBCD + mDAB = 360°
The circles is made up of arc BCD and arc DAB, therefore the sum angle of the arcs is equivalent to 360°
5. 2(m∠A + 2(m∠C) = 360; this is substitution property
From step 4 we stated that mBCD +mDAB = 360
but from the inscribed angle theorem;
mBCD= 2 (m∠A) and mDAB = 2(m∠C)
Therefore; substituting in the equation in step 4 we get;
2(m∠A) + 2(m∠C) = 360
idk sorry ill look for you
Answer:
Top left is 7x^2. Top right is 14x. Bottom left is -4x. Bottom right is -8.
Step-by-step explanation:
7x+10y=6400
x+y=700
x=700-y
7(700-y)+10y=6400
4900-7y+10y=6400
-7y+10y=1500
3y=1500
y=500
x=700-y
x=700-500
x=200
there were 200 children and 500 adults at the show
