Answer: 
Step-by-step explanation:
Since a Binomial is an algebraic expression of the sum or the difference of two terms.
Therefore, For a binomial an expression must contain exactly two terms.
In a)
there is only one term therefore it is not the Binomial.
In a)
, there is only one term therefore it is not the Binomial.
In a)
there is exactly two terms therefore it is a Binomial.
In a)
there are three terms therefore it is not the Binomial.
Answer:
1/10 = 2/20
1/3 = 3/9
5/7 = 15/21
Step-by-step explanation:
In order to figure this out you need to use
Descartes Rule. I attached a picture showing Descartes Rule. If the signs changes for when x is positive then the number of times it changes are the possible positive solution. If the sign changes when x is negative then the number of times it changes are the possible negative solutions. With that said the answer is A. View the picture I have attached for the possible + - and imaginary solutions.
Answer = A) One possible positive solution.
<span>x 2 3 4
f(x) 5.5 7 8.5 </span>
The first function is linear. When you subtract 5.5 from 7, you get 1.5 and when you subtract 7 from 8.5 you also get 1.5. That's how we know it's a linear function because there's clearly pattern. HOWEVER, if you had 5.5, 7, and 9 that would not be a linear function. Why? Because when you take away 7 from 9 you get 2. In order for it to be a linear function you have to get the same exact number when you subtract.
<span>x 0 3 6
f(x) 1 8 64</span>
The second function is exponential because when you divide you get the exact same number. 8/1 = 8 and 64/8 = 8. So this is exponential. But if you had 72 instead of 64 it would not be an exponential function because there has to be a pattern.
I hope this makes sense. Btw I had this question on my test too and this is correct.
Step-by-step explanation:
Hey there!
The equation of a circle which passes through any point having centre (h,k) is;

In first blank ( x-_ ) keep "-2". In second blank (y - _ ) keep "4" and in last blank ( _ ) keep "36".
<em><u>Hope</u></em><em><u> </u></em><em><u>it helps</u></em><em><u>.</u></em><em><u>.</u></em>