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1 year ago

Simplify the expression below. Then classify the resulting expression

Mathematics

2 answers:

maw [93]1 year ago

4 0

The answer is 5, which is a monomial.

Let's simplify using the distributive property.

3x² + 6x + 5 - 3x(2) - 3x(x)
3x² + 6x + 5 - 6x - 3x²
5

If the resulting expression has only one term, it is classified as a monomial.

evablogger [386]1 year ago

3 0

Answer:

Simplified = 5

Classification = Monomial

Step-by-step explanation:

<h2>PART I: Simplify the expression</h2>

Given expression:

3x² + 6x + 5 - 3x (2 + x)

Expand parenthesis by distributive property:

= 3x² + 6x + 5 - 3x (2) - 3x (x)

= 3x² +6x + 5 - 6x - 3x²

Put like terms together:

= 3x² - 3x² + 6x - 6x + 5

= 0 + 0 + 5

= $\boxed{5}$

<h2>PART II: Classify polynomial</h2>

Concept:

Polynomial is classified by the number of terms a polynomial has.

Monomial: a polynomial with only one term
Binomial: a polynomial with two terms
...

Classify the given expression:

Original = 3x² + 6x + 5 - 3x (2 + x)

Simplified = 5

5 is a constant and it has only one term

Therefore, it is a monomial.

Hope this helps!! :)

Please let me know if you have any questions

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Option a: The number of bacteria at time x is 0.

Option b: An exponential function that represents the population is $y=200(1.5)^x$

Option c: The population after 10 minutes is 11534(app)

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It is given that the coordinates of the graph are (0,200), (1,300) and (2, 450)

Option a: To determine the number of bacteria x when y = 200

From the graph, we can see that the line meets y = 200 when x = 0

Thus, the coordinates are (0,200)

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Option b: Now, we shall determine the exponential function of the population.

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To determine the common difference, let us divide,

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Thus, substituting the values $a=200$ and $b=1.5$ in the formula $y=a \cdot b^{x}$ ,

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Hence, An exponential function that represents the population is $y=200(1.5)^x$

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Thus, we have,

$\begin{aligned}y &=200(1.5)^{x} \\&=200(1.5)^{10} \\&=200(57.67) \\&=11534\end{aligned}$

Hence, the population after 10 minutes is 11534(app)

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