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3 years ago

For the polynomial 8x3y2 – x?y2 + 3xy2 – 4y3 to be fully simplified and written in standard form, the missing exponent on the

2 answers:

6 0

A polynomial in standard form has terms arranged with the highest exponent on leftmost side and the lowest exponent on the right end. The variables are also arranged alphabetically.

Therefore, the exponent of the x-term must be 2 for the polynomial to be fully simplified and written in standard form.

Karo-lina-s [1.5K]3 years ago

5 0

Answer: 2

Step-by-step explanation: this answer is 100% correct for edge.nuity

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Which statement is true about the end behavior of the function represented by the graph?

Zepler [3.9K]

End behavior always involves x approaching positive and negative infinity. So we'll cross off the choice that says "x approaches 1".

The graphs shows both endpoints going down forever. So both endpoints are going to negative infinity regardless if x goes to either infinity.

<h3>Answer: Choice B</h3><h3>As x approaches −∞, f(x) approaches −∞, and as x approaches ∞, f(x) approaches −∞.</h3>

Another way to phrase this would be to say "f(x) approaches negative infinity when x goes to either positive or negative infinity"

7 0

3 years ago

Geometry help please

Nana76 [90]

Answer:

CED = 117

Step-by-step explanation:

21x +21 = 7x +49

21x - 7x = 49 - 21

14x = 28

x = 28/14

x = 2

7x + 49

=7(2) + 49

=14 + 49

=63

180 -63 = 117

CED= 117

8 0

3 years ago

Read 2 more answers

Please help with the attached question. Thanks

Mademuasel [1]

Answer:

Choice A) $F(x) = 3\sqrt{x + 1}$ .

Step-by-step explanation:

What are the changes that would bring $G(x)$ to $F(x)$ ?

Translate $G(x)$ to the left by $1$ unit, and
Stretch $G(x)$ vertically (by a factor greater than $1$ .)

$G(x) = \sqrt{x}$ . The choices of $F(x)$ listed here are related to $G(x)$ :

Choice A) $F(x) = 3\;G(x+1)$ ;
Choice B) $F(x) = 3\;G(x-1)$ ;
Choice C) $F(x) = -3\;G(x+1)$ ;
Choice D) $F(x) = -3\;G(x-1)$ .

The expression in the braces (for example $x$ as in $G(x)$ ) is the independent variable.

To shift a function on a cartesian plane to the left by $a$ units, add $a$ to its independent variable. Think about how $(x-a)$ , which is to the left of $x$ , will yield the same function value.

Conversely, to shift a function on a cartesian plane to the right by $a$ units, subtract $a$ from its independent variable.

For example, $G(x+1)$ is $1$ unit to the left of $G(x)$ . Conversely, $G(x-1)$ is $1$ unit to the right of $G(x)$ . The new function is to the left of $G(x)$ . Meaning that $F(x)$ should should add $1$ to (rather than subtract $1$ from) the independent variable of $G(x)$ . That rules out choice B) and D).

Multiplying a function by a number that is greater than one will stretch its graph vertically.
Multiplying a function by a number that is between zero and one will compress its graph vertically.
Multiplying a function by a number that is between $-1$ and zero will flip its graph about the $x$ -axis. Doing so will also compress the graph vertically.
Multiplying a function by a number that is less than $-1$ will flip its graph about the $x$ -axis. Doing so will also stretch the graph vertically.

The graph of $G(x)$ is stretched vertically. However, similarly to the graph of this graph $G(x)$ , the graph of $F(x)$ increases as $x$ increases. In other words, the graph of $G(x)$ isn't flipped about the $x$ -axis. $G(x)$ should have been multiplied by a number that is greater than one. That rules out choice C) and D).

Overall, only choice A) meets the requirements.

Since the plot in the question also came with a couple of gridlines, see if the points $(x, y)$ 's that are on the graph of $F(x)$ fit into the expression $y = F(x) = 3\sqrt{x - 1}$ .