Answer: Question 1 :B,D.
Question 2:option B,
Question 3:Degree=5.
Question 4:option D.
Question 5: option c.
Step-by-step explanation:
1) A polynomial can not have any exponent as a variable or a fraction.
Options B and D are polynomials.
2) The polynomial is having 3 terms and is of degree 3.so it is a cubic trinomial Option B.
3) Degree is the highest power of the variables in the terms .The term 
has the power=3+2=5
So degree =5.
4)
Option D
5)
Simplifying like terms,
=
Option c.
The answer is B(11, 2sqrt(12) )
proof
the main equation of the circle is (x-x1)²+(y-y1)²=R²
where C(x1, y1) is the center
so if the center is the origin, it is O(0,0), and the equation becomes
<span> (x)²+(y)²=R²
</span>and the circle passes through the point (-5,2) so we can write
(-5)²+12²=R², it implies R= sqrt(25+144)=sqrt(169)=13
and for <span>B(11, 2sqrt(12) ) </span>11²+ (2sqrt(12))²= 121 + 48= 169= 13
it is checked.
Answer: 40% chance
Sample space: 3 white 2 red, 5 pairs
2/5 probability, as a percent would be 40% chance
Answer:
option A, option C, option D
Step-by-step explanation:
a) 1 ÷ m/6
can be written as
÷ 
b) sides in (m/6) will change if both has to multiply
c) 1 ÷ m/6
can be written as
1 * 6/m
1(
) and wont make change to answer. so matches with the question.
d)
1 ÷ m/6
1 * 6/m
1 * 6 * 
6 *
6 ÷ m ..therefore true
e)
1 ÷ m/6
1 * 6/m
6/m ....does not match or can be converted to the following so wrong
- Therefore A, C, D are correct and B and E is wrong.
Answer:
Lies in the shaded regions of both the top and bottom inequalities.
Step-by-step explanation:
The point of solution for BOTH systems of inequalities must work for both equations. Therefore, the point has to lie in both top and bottom shaded regions or it won't work for both, but just one.
