Given:
Quadrilateral ABCD is inscribed in a circle P.
To find:
Which statement is necessarily true.
Solution:
Quadrilateral ABCD is inscribed in a circle P.
Therefore ABCD is a cyclic quadrilateral.
In cyclic quadrilateral, opposite angles form a supplementary angles.
⇒ m∠A + m∠C = 180° --------- (1)
⇒ m∠B + m∠D = 180° --------- (2)
By (1) and (2),
⇒ m∠A + m∠C = m∠B + m∠D
This statement is necessarily true for the quadrilateral ABCD in circle P.
DC= 4
BD= 10
Hope it’s right
Best luck with your studying
First one :
10•5^2+4•-2^2=234
Second one :
-2(3•-2+-6)=24
<span>
−<span>2<span>(<span><span><span>(3)</span><span>(<span>−2</span>)</span></span>−6</span>)</span></span></span><span><span><span>
(<span>−2</span>)</span><span>(<span><span><span>(3)</span><span>(<span>−2</span>)</span></span>+<span>−6</span></span>)</span></span></span><span><span><span><span>
(<span>−2</span>)</span><span>(<span><span>(3)</span><span>(<span>−2</span>)</span></span>)</span></span>+<span><span>(<span>−2</span>)</span><span>(<span>−6</span>)</span></span></span></span><span>
<span>12+12</span></span><span><span>
24</span></span>
Last one :
2•-6(3•-6-5)
<span><span><span>
(2)</span><span>(<span>−6</span>)</span></span><span>(<span><span><span>(3)</span><span>(<span>−6</span>)</span></span>−5</span>)</span></span><span><span>
(<span><span>(2)</span><span>(<span>−6</span>)</span></span>)</span><span>(<span><span><span>(3)</span><span>(<span>−6</span>)</span></span>+<span>−5</span></span>)</span></span><span><span><span>
(<span><span>(2)</span><span>(<span>−6</span>)</span></span>)</span><span>(<span><span>(3)</span><span>(<span>−6</span>)</span></span>)</span></span>+<span><span>(<span><span>(2)</span><span>(<span>−6</span>)</span></span>)</span><span>(<span>−5</span>)</span></span></span><span>
216+60</span><span>
=276
Hoped I helped!</span>
If Im correct I believe the answer is the last one
