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kondor19780726 [428]
3 years ago
12

PLEASE HELP!!!! 20 POINTS!!!! What best describes the following equation?

Mathematics
1 answer:
TiliK225 [7]3 years ago
5 0
It’s both a relation and a function
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Solve:<br><br> a.x3+x2-8x - 12x+2<br> b. x3-4x2-3x + 18x-3<br> c. x2 + 4x + 4<br> d. x2 - 6x + 9
NeTakaya
<h2>Steps:</h2>

So firstly, this function is asking us to divide f(x) by g(x) so let's set that up as such:

(\frac{f}{g})(x)=\frac{x^2-x-6}{1}\div \frac{x-3}{x+2}

Next, remember that <u>dividing by a number is the same as multiplying by its reciprocal.</u> To find the reciprocal of a number, flip the numerator and denominator around. With this info, flip the second fraction to it's reciprocal and change the sign to multiplication:

(\frac{f}{g})(x)=\frac{x^2-x-6}{1}\times \frac{x+2}{x-3}

Next, we are going to factor x² - x - 6. Firstly, what two terms have a product of -6x² and a sum of -x? That would be -3x and 2x. Replace -x with 2x - 3x:

(\frac{f}{g})(x)=\frac{x^2+2x-3x-6}{1}\times \frac{x+2}{x-3}

Next, factor x² + 2x and -3x - 6 separately. Make sure that they have the same quantity on the inside of the parentheses:

(\frac{f}{g})(x)=\frac{x(x+2)-3(x+2)}{1}\times \frac{x+2}{x-3}

Now you can rewrite it as:

(\frac{f}{g})(x)=\frac{(x-3)(x+2)}{1}\times \frac{x+2}{x-3}

Next, multiply:

(\frac{f}{g})(x)=\frac{(x-3)(x+2)^2}{x-3}

Next, divide:

(\frac{f}{g})(x)=(x+2)^2

And lastly, simplify:

  • A good tip: (x+y)^2=x^2+2xy+y^2

(\frac{f}{g})(x)=x^2+4x+4

<h2>Answer:</h2>

<u>The correct option is C. x² + 4x + 4.</u>

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PLEASE HELP WITH GRADE 11 MATH. Make sure you show the formula. Substitute values and show all mathematicall operations! show yo
Degger [83]

3x+y

x

​

 

=−3

=−y+3

​

The second equation is solved for xxx, so we can substitute the expression -y+3−y+3minus, y, plus, 3 in for xxx in the first equation:

\begin{aligned} 3\blueD{x}+y &= -3\\\\ 3(\blueD{-y+3})+y&=-3\\\\ -3y+9+y&=-3\\\\ -2y&=-12\\\\ y&=6 \end{aligned}

3x+y

3(−y+3)+y

−3y+9+y

−2y

y

​

 

=−3

=−3

=−3

=−12

=6

​

Plugging this value back into one of our original equations, say x = -y +3x=−y+3x, equals, minus, y, plus, 3, we solve for the other variable:

\begin{aligned} x &= -\blueD{y} +3\\\\ x&=-(\blueD{6})+3\\\\ x&=-3 \end{aligned}

x

x

x

​

 

=−y+3

=−(6)+3

=−3

​

The solution to the system of equations is x=-3x=−3x, equals, minus, 3, y=6y=6y, equals, 6.

We can check our work by plugging these numbers back into the original equations. Let's try 3x+y = -33x+y=−33, x, plus, y, equals, minus, 3.

\begin{aligned} 3x+y &= -3\\\\ 3(-3)+6&\stackrel ?=-3\\\\ -9+6&\stackrel ?=-3\\\\ -3&=-3 \end{aligned}

3x+y

3(−3)+6

−9+6

−3

​

 

=−3

=

?

−3

=

?

−3

=−3

​

Yes, our solution checks out.

Example 2

We're asked to solve this system of equations:

\begin{aligned} 7x+10y &= 36\\\\ -2x+y&=9 \end{aligned}

7x+10y

−2x+y

​

 

=36

=9

​

In order to use the substitution method, we'll need to solve for either xxx or yyy in one of the equations. Let's solve for yyy in the second equation:

\begin{aligned} -2x+y&=9 \\\\ y&=2x+9 \end{aligned}

−2x+y

y

​

 

=9

=2x+9

​

Now we can substitute the expression 2x+92x+92, x, plus, 9 in for yyy in the first equation of our system:

\begin{aligned} 7x+10\blueD{y} &= 36\\\\ 7x+10\blueD{(2x+9)}&=36\\\\ 7x+20x+90&=36\\\\ 27x+90&=36\\\\ 3x+10&=4\\\\ 3x&=-6\\\\ x&=-2 \end{aligned}

7x+10y

7x+10(2x+9)

7x+20x+90

27x+90

3x+10

3x

x

​

 

=36

=36

=36

=36

=4

=−6

=−2

​

Plugging this value back into one of our original equations, say y=2x+9y=2x+9y, equals, 2, x, plus, 9, we solve for the other variable:

\begin{aligned} y&=2\blueD{x}+9\\\\ y&=2\blueD{(-2)}+9\\\\ y&=-4+9 \\\\ y&=5 \end{aligned}

y

y

y

y

​

 

=2x+9

=2(−2)+9

=−4+9

=5

​

The solution to the system of equations is x=-2x=−2x, equals, minus, 2, y=5y=5y, equals, 5.

5 0
2 years ago
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