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geniusboy [140]
3 years ago
12

Which statement shows how the product of (x + 7)2 demonstrates the closure property of multiplication?

Mathematics
2 answers:
Illusion [34]3 years ago
7 0

<h3>(x + 7)² = x² + 14x + 49</h3>

The true statement that show the closure property of multiplication :

<h3>a.  x² + 14x + 49 is a polynomial</h3>

\texttt{ }

<h3>Further explanation</h3>

<em>If equation ax³ + bx² + cx + d = 0 has roots x₁ , x₂ , and x₃ then</em>

x_1 + x_2 + x_3 = - \frac{b}{a}

x_1 x_2 + x_1 x_3 + x_2 x_3 = \frac{c}{a}

x_1 x_2 x_3 = - \frac{d}{a}

Let us now tackle the problem!

\texttt{ }

This problem is about Polynomial.

(x + 7)^2 = (x + 7)(x + 7)

(x + 7)^2 = x^2 + 7x + 7x + 7(7)

(x + 7)^2 = x^2 + 14x + 49

x^2 + 14x + 49 is a polynomial.

x + 7 is polynomial.

It demonstrates the closure property of multiplication.

\texttt{ }

<h3>Conclusion:</h3>

<em>The true statement that show how the product of (x + 7)² demonstrates the closure property of multiplication:</em>

<h3>a.  x² + 14x + 49 is a polynomial</h3>

\texttt{ }

<h3>Learn more</h3>
  • Solving Quadratic Equations by Factoring : brainly.com/question/12182022
  • Determine the Discriminant : brainly.com/question/4600943
  • Formula of Quadratic Equations : brainly.com/question/3776858

\texttt{ }

<h3>Answer details</h3>

Grade: High School

Subject: Mathematics

Chapter: Polynomial

\texttt{ }

Keywords: Quadratic , Equation , Discriminant , Real , Number

#LearnWithBrainly

Tasya [4]3 years ago
7 0

answer:

a.  x2 + 14x + 49 is a polynomial

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Step-by-step explanation:

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5 0
3 years ago
3. Solve the system using elimination (not substitution or matrices). negative 2 x plus y minus 2 z equals negative 8A N D7 x pl
riadik2000 [5.3K]

Elimination Method

\begin{gathered} -2X+Y-2Z=-8 \\ 7X+Y+Z=-1 \\ 5X+2Y-Z=-9 \end{gathered}

If we multiply the equation 3 by (-1) we obtain this:

\begin{gathered} -2X+Y-2Z=-8 \\ 7X+Y+Z=-1 \\ -5X-2Y+Z=9 \end{gathered}

If we add them we obtain 0, therefore there are infinite solutions. So, let's write it in terms of Z

1. Using the 3rd equation we can obtain X(Y,Z)

\begin{gathered} 5X=-9-2Y+Z \\ X=\frac{-9-2Y+Z}{5} \\  \end{gathered}

2. We can replace this value of X in the 1st and 2nd equations

\begin{gathered} -2\cdot(\frac{-9-2Y+Z}{5})+Y-2Z=-8 \\ 7\cdot(\frac{-9-2Y+Z}{5})+Y+Z=-1 \end{gathered}

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\begin{gathered} \frac{-9Y+12Z-63}{5}=-1 \\ \frac{9Y-12Z+18}{5}=-8 \end{gathered}

4. We can obtain Y from this two equations:

\begin{gathered} Y=-\frac{-12Z+58}{9} \\  \end{gathered}

5. Now, we need to obtain X(Z). We can replace Y in X(Y,Z)

\begin{gathered} X=\frac{-9-2Y+Z}{5} \\ X=\frac{-9-2(-\frac{-12Z+58}{9})+Z}{5} \end{gathered}

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X=\frac{-3Z+7}{9}

7. In conclusion, we obtain that

(X,Y,Z) =

(\frac{-3Z+7}{9},-\frac{-12Z+58}{9},Z)

8 0
1 year ago
02:42:33
jeyben [28]

Answer:    -10

Step-by-step explanation:

It is -10 because you have to switch the order of operations to get negative.

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3 years ago
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9514 1404 393

Answer:

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Step-by-step explanation:

<u>Given</u>:

  $1625 is invested at an annual rate of 1.95% compounded quarterly for 5 years

<u>Find</u>:

  the ending balance

<u>Solution</u>:

The compound interest formula applies.

  FV = P(1 +r/n)^(nt) . . . Principal P at rate r for t years, compounded n per year

  FV = $1625(1 +0.0195/4)^(4·5) = $1625(1.004875^20) ≈ $1790.99

The account ending balance would be $1790.99.

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