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Helen [10]
3 years ago
15

Please help me with this problem. I have tried many times and have gotten it wrong. I need help especially on the last part. Can

I also have a good explanation because I want to know how to do problems like this in the future. Also please answer at least the bottom two parts.

Mathematics
1 answer:
Arlecino [84]3 years ago
5 0

1) The domain is all the possible x values in the function so it would be [-4,4]

2) There are only 3 zeros shown on the graph and they are (-2, 0) (0, 0) (2, 0) the zeros are the value of x when y = 0.

3)The function is positive/Negative is asking for what x values make the y values positive aka interval notation. The function is positive if x = (0, 2) because 0 and 2 aren't included you use parentheses () instead of brackets []  

The function is negative if x [-4, -2), (-2,0), (2,4]

You might be interested in
Y = 3/5х + 2 on a graph
Andrei [34K]

Answer:

Go to desmos graphing calculator and put your equation in and it will graph it for you I used it for my math questions like this

Step-by-step explanation:

HOPE THAT HELPS YOU

7 0
2 years ago
Theorem: A line parallel to one side of a triangle divides the other two proportionately.
Kisachek [45]

Answer:

Segment BF = 16

Step-by-step explanation:

The given theorem states that a line parallel to one side of a triangle divides the other two sides proportionately

The given theorem is the Triangle Proportionality Theorem

According to the theorem, given that segment DE is parallel to segment BC, we have;

\dfrac{AD}{BD} = \dfrac{AE}{EC}

Therefore;

BD = \dfrac{AD}{\left(\dfrac{AE}{EC} \right) }  = AD \times \dfrac{EC}{AE}

Which gives;

BD = 6 \times \dfrac{18}{12}= 9

Similarly, given that EF is parallel to AB, we get;

\dfrac{AE}{EC} = \dfrac{BF}{FC}

Therefore;

BF = FC \times \dfrac{AE}{EC}

Which gives;

BF = 24 \times \dfrac{12}{18} = 16

Therefore, the statement that can be proved using the given theorem is segment BF = 16.

8 0
2 years ago
Which is the graph of y = kx] -2?
miss Akunina [59]

Please show the rest of the question if there is any

5 0
3 years ago
Guys, I really really need some help badly! I don't understand how to get the "GCF" aka greatest common factor. Can someone plea
Setler [38]

After I explain the procedure in full detail, you'll be able to generate
your own work.  Here it is.

To find the greatest common factor of two numbers:

1).  List all of the factors of the first number.

2).  List all of the factors of the second number.

3).  Scan both lists, and find factors that are on BOTH lists.
These are the COMMON factors.  Write them down in a short list.

4).  Scan the short list.  Find the biggest number on it.
That's the Greatest Common Factor.

OK, I'll go through the first one in your picture . . . 24  and  42 :

1). Factors of 24:  <em>1,  2,  3,</em>  4,<em>  6,</em>  8,  12,  24

2). Factors of 42:  <em>1,  2,  3,  6,</em>  7,  14,  21,  42

3). Short list:    <em>1, 2, 3, 6</em>

4). Biggest number on the short list:<em>  <u>6</u> </em>


5 0
3 years ago
Read 2 more answers
Will rate7 you brainliest
Vilka [71]

Answer:

\Large \boxed{\sf \bf \ \ \dfrac{x^2-x-6}{x^2-3x+2} \ \ }

Step-by-step explanation:

Hello, first of all, we will check if we can factorise the polynomials.

\boxed{x^2+6x+8}\\\\\text{The sum of the zeroes is -6=(-4)+(-2) and the product 8=(-4)*(-2), so}\\\\x^2+6x+8=x^2+2x+4x+8=x(x+2)+4(x+2)=(x+2)(x+4)

\boxed{x^2+3x-10}\\\\\text{The sum of the zeroes is -3=(-5)+(+2) and the product -10=(-5)*(+2), so}\\\\x^2+3x-10=x^2+5x-2x-10=x(x+5)-2(x+5)=(x+5)(x-2)

\boxed{x^2+2x-15}\\\\\text{The sum of the zeroes is -2=(-5)+(+3) and the product -15=(-5)*(+3), so}\\\\x^2+2x-15=x^2-3x+5x-15=x(x-3)+5(x-3)=(x+5)(x-3)

\boxed{x^2+3x-4}\\\\\text{The sum of the zeroes is -3=(-4)+(+1) and the product -4=(-4)*(+1), so}\\\\x^2+3x-4=x^2-x+4x-4=x(x-1)+4(x-1)=(x+4)(x-1)

Now, let's compute the product.

\dfrac{x^2+6x+8}{x^2+3x-10}\cdot \dfrac{x^2+2x-15}{x^2+3x-4}\\\\\\=\dfrac{(x+2)(x+4)}{(x+5)(x-2)}\cdot \dfrac{(x+5)(x-3)}{(x+4)(x-1)}\\\\\\\text{We can simplify}\\\\=\dfrac{(x+2)}{(x-2)}\cdot \dfrac{(x-3)}{(x-1)}\\\\\\=\large \boxed{\dfrac{x^2-x-6}{x^2-3x+2}}

So the correct answer is the first one.

Thank you.

7 0
3 years ago
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