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2 years ago

What is limit of (x cubed 3 x squared 4 x minus 12) as x approaches negative 3? –78 –24 30 54

Mathematics

1 answer:

Sergio039 [100]2 years ago

7 0

The limit of the expression as x approaches -3 is -24

<h3>How to determine the limit of the expression?</h3>

The expression is given as:

$x^3 + 3x^2 + 4x - 12$

As x approaches -3.

The limit expression becomes

$\lim_{x \to -3} x^3 + 3x^2 + 4x - 12$

Substitute -3 for x in the expression

$\lim_{x \to -3} (-3)^3 + 3(-3)^2 - 4*3 - 12$

Evaluate the expression

$\lim_{x \to -3} -24$

Hence, the limit of the expression as x approaches -3 is -24

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On a coordinate plane, a line goes through (0, negative 1) and (3, 1). a point is at (negative 3, 0). what is the equation of th

Vitek1552 [10]

Required Equation is $y = \frac{2x}{3}+2$

Given: A line passing through $({x}_1,{y}_1)$ and $({x}_2,{y}_2)$

Slope = $\frac{(y_2 - y_1)}{(x_2 - x_1)}$ = $\frac{( 1 - (-1))}{(3 - 0)}= \frac{2}{3}$

Equation of line is:

$Y_1(x) = \frac{2}{3}x + b$

Where b is y intercept

for value of b

$Y_1(0) = -1 = \frac{2}{3}*0 + b = b$

⇒ b = -1

and Equation of line is:

$Y_1(x) = \frac{2}{3}x + (-1)$

For another parallel line that passes through the point (-3,0)

$Y_2(x)= \frac{2}{3}x + c$

$Y_2(-3) = 0 = {2}{3}*-3 + c = -2 + c = 0$

c = 2

then the equation is:

$Y_2(x) = \frac{2}{3}x + 2$

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5 0

2 years ago

Why does he look so off

victus00 [196]

He’s a cocain addict. They snort them Poland bro

8 0

2 years ago

Read 2 more answers

Calculate the sample standard deviation and sample variance for the following frequency distribution of hourly wages for a sampl

lidiya [134]

Answer:

(a) The sample variance is 16.51

(a) The sample standard deviation is 4.06

Step-by-step explanation:

Given

$\begin{array}{cc}{Class} & {Frequency} & 8.26 - 10.00 & 20 &10.01-11.75 & 38 &11.76 - 13.50& 36 & 13.51-15.25 &25&15.26-17.00 &27 &\ \end{array}$

Solving (a); The sample variance.

First, calculate the class midpoints.

This is the mean of the intervals.

i.e.

$x_1 = \frac{8.26+10.00}{2} = \frac{18.26}{2} = 9.13$

$x_2 = \frac{10.01+11.75}{2} = \frac{21.76}{2} = 10.88$

$x_3 = \frac{11.76+13.50}{2} = \frac{25.26}{2} = 12.63$

$x_4 = \frac{13.51+15.25}{2} = \frac{28.76}{2} = 14.38$

$x_5 = \frac{15.26+17.00}{2} = \frac{32.26}{2} = 16.13$

So, the table becomes:

$\begin{array}{ccc}{Class} & {Frequency} & {x} & 8.26 - 10.00 & 20&9.13 &10.01-11.75 & 38 &10.88&11.76 - 13.50& 36 &12.63& 13.51-15.25 &25&14.38&15.26-17.00 &27 &16.13\ \end{array}$

Next, calculate the mean

$\bar x = \frac{\sum fx}{\sum f}$

$\bar x = \frac{20*9.13 + 38 * 10.88+36*12.63+25*14.38+27*16.13}{20+38+36+25+27}$

$\bar x = \frac{1845.73}{146}$

$\bar x = 12.64$

Next, the sample variance is:

$\sigma^2 = \frac{\sum f(x - \bar x)^2}{\sum f - 1}$

So, we have:

$\sigma^2 = \frac{20*(9.13-12.63)^2 + 38 * (10.88-12.63)^2 +...........+27 * (16.13 -12.63)^2}{20+38+36+25+27-1}$

$\sigma^2 = \frac{2393.6875}{145}$

$\sigma^2 = 16.51$

The sample standard deviation is:

$\sigma = \sqrt{\sigma^2}$

$\sigma = \sqrt{16.51}$

$\sigma = 4.06$

6 0

3 years ago

My homework says, "Does the graph represent a proportional relationship? Why or why not? You MUST explain to get credit."

VMariaS [17]

Answer:

No, the graph does not represent a proportional relationship. It is a straight line, but it does not intersect the point (0, 0). ... It's not proportional because the line does not cross through the origin. Therefore the ratio of x to y will be different for two different points on the line.

4 0

3 years ago

In January Jennifer was allowed to use 150 text messages. If she is allowed to use the same amount of text messages every month.

Firlakuza [10]

She would have used 1200 messages in all by the month of August.

Step-by-step explanation:

Messages allowed per month = 150

There will be 8 months starting from January to August, therefore,

Time period = 8 months

Messages in 8 months = Messages per month * Time period

$Messages\ in\ 8\ months=150*8\\Messages\ in\ 8\ months=1200$

She would have used 1200 messages in all by the month of August.

Keywords: Multiplication

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6 0

3 years ago