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goldfiish [28.3K]
3 years ago
8

Gina Wilson, all things algebra unit 5 homework 2

Mathematics
1 answer:
Vilka [71]3 years ago
5 0

We have that for the Question,it can be said that these the various <em>graphs</em> and polynomials have the following deductions

1)

Even degree

<em>Negative </em>leading <em>coefficient</em>

  • Second graph

2)

Odd degree

Positive leading <em>coefficient</em>

  • 1st Graph

3)

The end behaviour of the 14th diploma <em>polynomial </em>is that it will increase to infinity.

4)

The <u>polynomial</u> will have a tendency to infinity.

Generally

The end behavior of a <em>polynomial </em><u>gra</u>ph draws reference from the starting <em>direction </em>and its end direction or the <em>ends </em>of the x axis

Where

Graph 1

f(x)= -\infty (Left)\\\\f(x)= +\infty (Right)

A Graph of even or odd degree bears the following lead co-efficient characteristics

Even

f(x) -> \infty \ as x -> \pm \infty  \\\\f(x) -> -\infty \ as x -> \pm \infty

Odd

f(x) -> -\infty \as x -> - \infty\\\\f(x) -> \infty \ as x ->  \infty

Therefore

  • 1st Graph

Positive leading <em>coefficient</em>

Odd degree

  • Second graph

<em>Negative </em>leading <em>coefficient</em>

Even degree

3)

Even Numbered degree <u>typically </u>have the <em>identical</em> give up behavior for the two ends. This his due to the fact that if N is a entire number,

-A^2=A^2  

Due to the fact the Leading <em>coefficient </em>is positive, and a variety with an even exponent is additionally positive,  end behaviour of the 14th diploma <em>polynomial </em>is that it will increase to infinity.

4)

The ninth degree polynomial  as we have a leading <em>coefficient </em>and a abnormal exponent.  

Then as x tends to infinity, the polynomial will have a <em>tendency</em> to terrible infinity. as x tends to -ve<u> </u>infinity, the <u>polynomial</u> will have a tendency to infinity.

For more information on this visit

brainly.com/question/12021944

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Answer:

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Step-by-step explanation:

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3 years ago
Y^2+X^2=36 y=-3x+5.
lesya [120]

Answer:

The two solutions in exact form are:

(\frac{15+\sqrt{335}}{10},\frac{5-3\sqrt{335}}{10})

and

(\frac{15-\sqrt{335}}{10},\frac{5+3\sqrt{335}}{10}).

If you prefer to look at approximations just put into your calculator:

(3.3303,-4.9909)

and

(-0.3303,5.9909).

Step-by-step explanation:

I guess you are asked to find the solution the given system.

I'm going to use substitution.

This means I'm going to plug the second equation into the first giving me:

(-3x+5)^2+x^2=36  I replaced the 1st y with what the 2nd y equaled.

Before we continue solving this I'm going to expand the (-3x+5)^2 using the following:

(a+b)^2=a^2+2ab+b^2.

(-3x+5)^2=(-3x)^2+2(-3x)(5)+(5)^2

(-3x+5)^2=9x^2-30x+25

Let's go back to the equation we had:

(-3x+5)^2+x^2=36  

After expansion of the squared binomial we have:

9x^2-30x+25+x^2=36

Combine like terms (doing the 9x^2+x^2 part:

10x^2-30x+25=36

Subtract 36 on both sides:

10x^2-30x+25-36=0

Simplify the 25-36 part:

10x^2-30x-11=0

Compare this to ax^2+bx+c=0 which is standard form for a quadratic.

We should see the following:

a=10

b=-30

c=-11

The formula that solves this equation for the variable x is:

x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}

Plugging in our values for a,b, \text{ and } c give us:

x=\frac{30 \pm \sqrt{(-30)^2-4(10)(-11)}}{2(10)}

Simplify the bottom; that is 2(10)=20:

x=\frac{30 \pm \sqrt{(-30)^2-4(10)(-11)}}{20}

Put the inside of square root into the calculator; that is put (-30)^2-4(10)(-11) in the calculator:

x=\frac{30 \pm \sqrt{1340}}{20}

Side notes before continuation:

Let's see if 1340 has a perfect square.

I know 1340 is divisible by 10 because it ends in 0.

1340=10(134)

134 is even so it is divisible by 2:

1340=10(2)(67)

1340=2(2)(5)(67)

1340=4(5)(67)

1340=4(335)

4 is a perfect square so we can simplify the square root part further:

\sqrt{1340}=\sqrt{4}\sqrt{335}=2\sqrt{335}.

Let's go back to the solution:

x=\frac{30 \pm \sqrt{1340}}{20}

x=\frac{30 \pm 2 \sqrt{335}}{20}

Now I see all three terms contain a common factor of 2 so I'm going to divide top and bottom by 2:

x=\frac{\frac{30}{2} \pm \frac{2 \sqrt{335}}{2}}{\frac{20}{2}}

x=\frac{15 \pm \sqrt{335}}{10}

So we have these two x values:

x=\frac{15+\sqrt{335}}{10} \text{ or } \frac{15-\sqrt{335}}{10}

Now we just need to find the corresponding y-coordinate for each pair of points.

I'm going to use the easier equation y=-3x+5.

Let's do it for the first x I mentioned:

If x=\frac{15+\sqrt{335}}{10} then

y=-3(\frac{15+\sqrt{335}}{10})+5.

Let's simplify:

Distribute the -3 to the terms on top:

y=\frac{-45-3\sqrt{335}}{10}+5

Combine the two terms; I'm going to do this by writing 5 as 50/10:

y=\frac{-45-3\sqrt{335}+50}{10}

Combine like terms on top; the -45+50 part:

y=\frac{5-3\sqrt{335}}{10}.

So one solution point is:

(\frac{15+\sqrt{335}}{10},\frac{5-3\sqrt{335}}{10}).

Let's find the other one for the other x that we got.

If x=\frac{15-\sqrt{335}}{10} then

y=-3(\frac{15-\sqrt{335}}{10})+5.

Let's simplify.

Distribute the -3 on top:

y=\frac{-45+3\sqrt{335}}{10}+5

I'm going to write 5 as 50/10 so I can combine the terms as one fraction:

y=\frac{-45+3\sqrt{335}+50}{5}

Simplify the -45+50 part:

y=\frac{5+3\sqrt{335}}{10}.

So the other point of intersection is:

(\frac{15-\sqrt{335}}{10},\frac{5+3\sqrt{335}}{10}).

The two solutions in exact form are:

(\frac{15+\sqrt{335}}{10},\frac{5-3\sqrt{335}}{10})

and

(\frac{15-\sqrt{335}}{10},\frac{5+3\sqrt{335}}{10}).

If you prefer to look at approximations just put into your calculator:

(3.3303,-4.9909)

and

(-0.3303,5.9909).

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3 years ago
Carlos had48 brownies. He ate 3 brownies and then gave 2 brownies to each of his 16 friends. How many brownies does Carlos have
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Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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3 years ago
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