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Agata [3.3K]
3 years ago
14

Have you done any factoring with variables? Which property (Commutative, Distributive, or Associative) do you think is most invo

lved in factoring? Do you think a polynomial can be a factor of another polynomial? Answer in full sentences.
Mathematics
2 answers:
vazorg [7]3 years ago
7 0
Idk how im supposed to answer this or whatevs bu
yes i have done factoring with variables.
distrubutive is the most involved in factoring method as i can tell.
i dont think ive studied this before or i mightve forgoten but im sure a polynomial can be factored into another one.
and yea. hope this helped in sum way
ASHA 777 [7]3 years ago
3 0

⇒x²y³=x*x*y*y*y=x*y*x*y*y=y*x*y*x*y--Writing factors in another way

⇒xy²z²=x*y*y*z*z=y*x*z*y*z=z*x*y*x*y*z---Writing factors in another way.

⇒xy+xy²=xy(1+y)

⇒y²z²-yzx=yz(yz-x)

Here ,we have used law of indices first, then used Commutativity , Associativity and Distributivity to factor Polynomial.

1.\rightarrow x^{a+b}=x^a \times x^b\\\\2.\rightarrow x \times y=y \times x----\text{Commutative Property}\\\\3.\rightarrow x \times (y \times z)= (x \times y) \times z=(x \times z) \times y---\text{Associaitive Property}

⇒Distributive Property of Multiplication with respect to addition and Subtraction

  →a×(b+c)=a×b+a×c

 → a×(b-c)=a×b-a×c

⇒Yes, A polynomial can be factor of another polynomial.

For Example:

⇒ x²-a², has two factors equal to , (x+a) and (x-a), where , x²-a², is a quadratic polynomial and two of it's factors are ,(x+a) and (x-a) which is also a polynomial of degree 1.

 

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On the last three physics exams a student scored 89, 87, and 92. What score must the student earn on the next exam to have an av
OlgaM077 [116]

Answer:

For the average to be above 90, the marks in the fourth paper should be MORE THAN 92.

Step-by-step explanation:

The score of students in last 3 exams =  89 , 87 and 92

Let us assume the marks scored by the student in last exam = k

The average of marks should be above 90.

Now, \textrm{The average of total marks}  = \frac{\textrm{Sum of all marks obtained in 4 papers}}{\textrm{ 4 papers }}

Sum of all 4 paper's marks =  89 + 87 + 92 + k = 268 + k

Also, 90 <  Average

\implies 90 < \frac{268 + k}{4}\\ or, 360 < 268 + k\\\implies  360 - 268 < k\\or,  k >  92

Hence,for the average to be above 90, the marks in the fourth paper should be MORE THAN 92.

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