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3 years ago

Simplify the expression: 3x + 5x + 6x – 4

Mathematics

2 answers:

geniusboy [140]3 years ago

7 0

Answer:

14x-4

Step-by-step explanation:

So you know 3x +5x = 8x so then you add 6x and put -4

77julia77 [94]3 years ago

5 0

Answer:

14x - 4

Step-by-step explanation:

simplifying this kind of given expression just means to put the variables all together and the numbers all together to get a shorter and more simplified version of the equation.

first do:

3x+5x+6x=

14x

and than you just have one number so your final simplified equation would be:

14x-4

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A) The equilibrium prices P1 and P2 for two goods satisfy the equations:

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The value of P1 and P2 using the inverse matrix is 5 and 6 respectively.

<h3 />

The equilibrium prices of the three independent commodities using the inverse matrix are P1 = 4, P2 = 7, P3 = 8 respectively.

<h3>What is a matrix?</h3>

A matrix can be defined as a collection of integers(numbers that are either positive or negative) that are organized in rows and columns to construct a rectangular array. The numbers in this matrix system are referred to as elements.

To determine the values of P1 and P2 for the system of equations given by using an inverse matrix, we have:

9P1 + P2 = 51

3P1 +4P2 = 39

Representing the above data in matrix form, we have:

$\left[\begin{array}{cc}9&1\\3&4\\ \end{array}\right] \left[\begin{array}{c} \mathbf{P_1} \\ \mathbf{P_2}\\ \end{array}\right] = \left[\begin{array}{c} \mathbf{51} \\ \mathbf{39} \\ \end{array}\right]$ which is in the form AX = B

In order for us to determine the values of P1 and P2, Let take the inverse of A⁻¹ on both sides of the AX= B, we have:

$\mathbf{{A^{-1} AX} = A^{-1} B}$

X = A⁻¹ B

Let's start by finding A⁻¹;

$\mathbf{A = \left[\begin{array}{cc}9&1\\3&4\end{array}\right] }$

$\mathbf{A^{-1} = \dfrac{1}{36-3}\left[\begin{array}{cc}4&-1\\3&9\end{array}\right] }$

$\mathbf{A^{-1} = \dfrac{1}{33}\left[\begin{array}{cc}4&-1\\3&9\end{array}\right] }$

Now, Let's Find A⁻¹B;

$\mathbf{A^{-1}B = \dfrac{1}{33}\left[\begin{array}{cc}4&-1\\3&9\end{array}\right] \left[\begin{array}{c}51\\39\\ \end{array}\right] }$

$\mathbf{\implies \dfrac{1}{33}\left[\begin{array}{cc}204&-39\\-153&+351\end{array}\right] }$

$\mathbf{\implies \left[\begin{array}{c}\dfrac{165}{33}\\ \\ \dfrac{198}{33}\end{array}\right] }$

$\mathbf{\implies \left[\begin{array}{c}5\\ \\ 6\end{array}\right] }$

$\left[\begin{array}{c}\mathbf{P_1}\\ \mathbf{P_2}\end{array}\right]= \left[\begin{array}{c}5\\ 6 \end{array}\right] }$

Therefore, we can conclude that the value of P1 and P2 using the inverse matrix is 5 and 6 respectively.

To determine the equilibrium prices of the three independent commodities using the inverse matrix, we have:

P₁ + 2P₂ + 3P₃ = 42

2P₁ + P₂ + 4P₃ = 47

P₁ + 3P₂ + 4P₃ = 57

The matrix in AX = B form is computed as:

$\implies\left[ \begin{array}{ccc}1&2&3\\2&1&4\\1&3&4\end{array}\right] \left[\begin{array}{c}P_1\\P_2\\P_3 \end{array}\right] = \left[\begin{array}{c}42\\47\\ 57\end{array}\right]$

$\mathbf{A^{-1} = \dfrac{1}{|A|} \ (adj \ A)}$

$\mathbf{A^{-1} = \dfrac{1}{1(4-12) -2(8-4) +3(6-1)} \left[\begin{array}{ccc}-8&4&5\\1&1&-1\\5&2&-3\end{array}\right] }^1$

$\mathbf{A^{-1}B = -1 \left[\begin{array}{ccc}-8&1&5\\-4&1&2\\5&2&-3\end{array}\right] }\left[\begin{array}{c}42\\47\\57\end{array}\right]$

$\mathbf{A^{-1}B = -\left[\begin{array}{ccc}-336&+47&+285\\-168&+47&+114\\210&-47&-171\end{array}\right] }$

$\mathbf{A^{-1}B = -\left[\begin{array}{c}-4\\-7\\-8\end{array}\right] }$

$\mathbf{A^{-1}B = \left[\begin{array}{c}4\\7\\8\end{array}\right] }$

Therefore, we can conclude that the values of P1 = 4, P2 = 7, P3 = 8 respectively.

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$\mathrm{Convert\:mixed\:numbers\:to\:improper\:fractions}:\quad 2\frac{1}{2}=\frac{5}{2}\\=\frac{5}{2}\div \:3\frac{2}{3}\\\mathrm{Convert\:mixed\:numbers\:to\:improper\:fractions}:\quad 3\frac{2}{3}=\frac{11}{3}\\=\frac{5}{2}\div \frac{11}{3}\\\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{a}{b}\div \frac{c}{d}=\frac{a}{b}\times \frac{d}{c}\\=\frac{5}{2}\times \frac{3}{11}\\\mathrm{Multiply\:fractions}:\quad \frac{a}{b}\times \frac{c}{d}=\frac{a\:\times \:c}{b\:\times \:d}\\$

$\mathrm{Multiply\:fractions}:\quad \frac{a}{b}\times \frac{c}{d}=\frac{a\:\times \:c}{b\:\times \:d}\\=\frac{5\times \:3}{2\times \:11}\\\mathrm{Multiply\:the\:numbers:}\:5\times \:3=15\\=\frac{15}{2\times \:11}\\\mathrm{Multiply\:the\:numbers:}\:2\times \:11=22\\=\frac{15}{22}$

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3 years ago

Will give Brainliest (:

olya-2409 [2.1K]

Answer:

Step-by-step explanation:

2x^2+8x-12=0

with form

ax^2+bx+c=0

move c to the other side

2x^2+8x=12

x^2 is multiplied -> divide equation by 2

x^2+4x=6

complete square:

divide b=4 by 2

4*1/2=2

square 2

2^2=4

add 4 to both sides

x^2+4x=6

x^2+4x+4=6+4

x^2+4x+4=10

transform to polynom

(x+2)^2=10

this left side is option B multiplied by a factor of 16 and therefore equal:

(4x + 8)^2

(4*4)x^2+(4*2*8)x+8*8

16x^2+64x+64

x^2+4+4

bonus: calculate root

x+2=+/-sqrt(10)

x=-2+/-sqrt(10)

<h2 /><h2>so it is option B</h2>

<em>2x^2 + 8x - 12 = 0..divide by 2</em>
<em>x^2 + 4x - 6 = 0</em>
<em>x^2 + 4x = 6...add 4 to both sides of the equation</em>
<em>x^2 + 4x + 4 = 6 + 4</em>
<em>(x + 2)^2 = 10....<== ur constant is 10</em>
<em>x + 2 = (+-)sqrt 10</em>
<em>x = -2 (+ - ) sqrt 10</em>
<em></em>
<em>x = -2 + sqrt 10</em>
<em>x = -2 - sqrt 10</em>
<em></em>
<em></em>
<em></em>

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3 years ago

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