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.
(x² + 3x - 1)(2x² - 2x + 1)
x²(2x² - 2x + 1) + 3x(2x² -2x + 1) -1(2x² - 2x + 1)
2x^4 - 2x³ + x² + 6x³ - 6x² + 3x - 2x² + 2x - 1
2x^4 - 2x³ + 6x³ + x² - 6x² - 2x² + 3x + 2x - 1
2x^4 + 4x³ - 7x² + 5x - 1
<span>(D)The result 2x4 + 4x3 − 7x2 + 5x − 1 is a polynomial.</span>
HE will need to refill his tank because on his return he will be 6 minles shrt of being back.
9514 1404 393
Answer:
D.) a+2b
Step-by-step explanation:
The integers 'a' and 'b' can be any, so you can choose a couple and evaluate these expressions to see what you get. For example, we can let a=1 and b=0. For these values, the offered expressions evaluate to ...
A) 3(0) = 0 . . . even
B) 1 +3 = 4 . . . even
C) 2(1+0) = 2 . . . even
D) 1 +2(0) = 1 . . . odd
_____
<em>Additional comment</em>
These rules apply to even/odd:
- odd × odd = odd
- odd × even = even
- even × even = even
- odd + odd = even
- odd + even = odd
- even + even = even
Then A is (odd)(even) = even; B is (odd)+(odd) = even; C is (even)(whatever) = even; D = (odd)+(even) = odd.
