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artcher [175]
4 years ago
8

How are the commutative property of addition and the commutative property of multiplication alike?

Mathematics
1 answer:
Rasek [7]4 years ago
6 0
Both condone the swapping of terms being operated on through multiplication or addition. For example: 

4 * 6 = 24 OR 6 * 4 = 24

7+3 = 10 OR 3 + 7 = 10

Commutative comes from the same root wrd as commute, so just think of it meaning "to move around".
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108 is 36% of what number? Write and solve a proportion to solve the problem
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3 years ago
Find a function of the form y=ax^2+bx+c whose graph passes through (-11,2),(-9,-2)and (-5,14). (Vertex form also?) please help
nikitadnepr [17]

Answer:

y=x^2+18x+79

y=(x+9)^2-2

Step-by-step explanation:

We want to find a function of the form:

y=ax^2+bx+c

We know that it passes through the three points: (-11, 2); (-9, -2); and (-5,14).

In other words, if we substitute -11 for <em>x</em>, we should get 2 for <em>y</em>. So:

(2)=a(-11)^2+b(-11)+c

Simplify:

2=121a-11b+c

We will do it to the two other points as well. So, for (-9, -2):

\begin{aligned}(-2)&=a(-9)^2+b(-9)+c\\\Rightarrow-2&=81a-9b+c\end{aligned}

And similarly:

\begin{aligned}(14)&=a(-5)^2+b(-5)+c\\\Rightarrow 14&=25a-5b+c\end{aligned}

So, to find our function, we will need to determine the values of a, b, and c.

We essentially have a triple system of equations:

\begin{cases}2=121a-11b+c\\-2=81a-9b+c\\14=25a-5b+c\end{cases}

To approach this, we can first whittle it down only using two variables.

So, let's use the First and Second Equation. Let's remove the variable <em>b</em>. To do so, we can use elimination. We can multiply the First Equation by 9 and the Second Equation by -11. This will yield:

\begin{cases}{\begin{aligned}9(2)&=9(121a-11b+c)\\-11(-2)&=-11(81a-9b+c)\end{cases}}

Distribute:

\begin{cases}18=1089a-99b+9c\\22=-891a+99b-11c\end{cases}

Now, we can add the two equations together:

(18+22)=(1089a-891a)+(-99b+99b)+(9c-11c)

Simplify:

40=198a-2c

Now, let's do the same using the First and Third Equations. We want to cancel the variable <em>b</em>. So, let's multiply the First Equation by 5 and the Third Equation by -11. So:

\begin{cases}{\begin{aligned}5(2)=5(121a-11b+c)\\-11(14)=-11(25a-5b+c)\end{cases}

Simplify:

\begin{cases}10=605a-55b+5c\\-154=-275a+55b-11c\end{cases}

Now, let's add the two equations together:

(10-154)=(605a-275a)+(-55b+55b)+(5c-11c)

Simplify:

-144=330a-6c

Therefore, we now have the two equations:

\begin{cases}40=198a-2c\\-144=330a-6c\end

Let's cancel the <em>c</em>. So, multiply the First Equation by -3. We don't have to do anything special to the second. So:

-3(40)=-3(198a-2c)

Multiply:

-120=-594a+6c

Now, add it to the Second Equation:

(-120+-144)=(-594a+330a)+(6c-6c)

Add:

-264=-264a

Divide both sides by -264. So, the value of <em>a</em> is:

a=1

Now, we can use either of the two equations above to obtain <em>c</em>. Let's use the first one. So:

40=198a-2c

Substitute 1 for <em>a: </em>

<em />40=198(1)-2c<em />

Solve for <em>c</em>. Subtract 198 from both sides and divide by -2:

<em />\begin{aligned}40&=198-2c\\ -158&=-2c \\79 &=c \end{aligned}<em />

So, the value of <em>c</em> is 79.

<em />

Finally, we can find b. We can use any of the three original equations. Let's use the First Equation. So:

<em />2=121a-11b+c<em />

Substitute 1 for <em>a</em> and 79 for <em>c</em> and determine the value of <em>b: </em>

<em />\begin{aligned}2&=121(1)-11b+(79)\\2&=121+79-11b\\2&=200-11b\\-198&=-11b\\18&=b\end{cases}<em />

Therefore, the value of <em>b</em> is 18.

So, our equation is:

<em> </em>y=ax^2+bx+c<em />

Substitute in the values:

y=(1)x^2+(18)x+(79)

Simplify:

y=x^2+18x+79

Now, let's put this into vertex form. To do so, we will need to complete the square. First, let's group the first two terms together:

y=(x^2+18x)+79

To complete the square, we will divide the <em>b</em> term by 2 and then square it.

<em>b</em> is 18. 18/2 is 9 and 9² is 81. Therefore, we will add 81 into our parentheses:

y=(x^2+18x+81)+79

Since we added 81, we must also subtract 81. So:

y=(x^2+18x+81)+79-81

Subtract:

y=(x^2+18x+81)-2

The grouped terms are a perfect square trinomial. Factor:

y=(x+9)^2-2

And this is in vertex form.

And we are done!

6 0
3 years ago
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