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mestny [16]
3 years ago
15

A rectangle has a length of (6x+1) units and a width of (3x+1 units. Express the area of the rectangle as trinomial. Please I re

ally need help and my tutor didn’t help. Please explain the work u did
Mathematics
1 answer:
dimulka [17.4K]3 years ago
8 0

whenever you have say a multiplication of a binomial or any polynomial, you can simply multiply each term of one by the other's terms, namely

(a+b)*(c+d+e)  => a(c+d+e) + b(c+d+e), and then add like-terms.

\bf \stackrel{length}{(6x+1)}\stackrel{width}{(3x+1)}\implies 6x(3x+1)+1(3x+1)\implies (18x^2+6x)~~+~~(3x+1) \\\\\\ 18x^2+6x+3x+1\implies \stackrel{\textit{adding like-terms}}{18x^2+9x+1}

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What is the vertex of the graph of f(x) = x2 + 10x - 9? A) (-5, -34) B) (-5, -9) C) (5, -9) D) (5, 66)
sukhopar [10]

Answer:

A) (-5, -34)

Step-by-step explanation:

f(x) = x^2 + 10x - 9

We complete the square to get the equation in vertex form

Take the coefficient of the x term and divide by 2 then square it.  We add it and then subtract it not to change the value of the equation

f(x) = x^2 + 10x +(10/2)^2 - (10/2)^2 - 9

f(x) = x^2 +10x +25 -25 -9

f(x) = (x^2 +10x +25) -34

The term in parentheses simplified to (x+10/2) ^2

     = (x+5)^2 -34

   = (x - -5)^2 -34

This is in the form (x-h)^2 +k

The vertex is (h,k)  h=-5 and k=-34

(-5,-34)

4 0
2 years ago
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Let P and Q be polynomials with positive coefficients. Consider the limit below. lim x→[infinity] P(x) Q(x) (a) Find the limit i
jenyasd209 [6]

Answer:

If the limit that you want to find is \lim_{x\to \infty}\dfrac{P(x)}{Q(x)} then you can use the following proof.

Step-by-step explanation:

Let P(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots+a_{1}x+a_{0} and Q(x)=b_{m}x^{m}+b_{m-1}x^{n-1}+\cdots+b_{1}x+b_{0} be the given polinomials. Then

\dfrac{P(x)}{Q(x)}=\dfrac{x^{n}(a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)}+a_{0}x^{-n})}{x^{m}(b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m})}=x^{n-m}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)})+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}

Observe that

\lim_{x\to \infty}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)})+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}=\dfrac{a_{n}}{b_{m}}

and

\lim_{x\to \infty} x^{n-m}=\begin{cases}0& \text{if}\,\, nm\end{cases}

Then

\lim_{x\to \infty}=\lim_{x\to \infty}x^{n-m}\dfrac{a_{n}+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_{2}x^{-(n-2)}+a_{1}x^{-(n-1)}+a_{0}x^{-n}}{b_{m}+b_{m-1}x^{-1}+b_{n-2}x^{-2}+\cdots+b_{2}x^{-(m-2)}+b_{1}x^{-(m-1)}+b_{0}x^{-m}}=\begin{cases}0 & \text{if}\,\, nm \end{cases}

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Answer:

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Step-by-step explanation:

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