Select all the factors of x^3+5x^2+2x-8

Please help! I will mark you as brainliest!

matrenka [14]

Answer:

3.5 hours traveling

2.5 hours stationary

Step-by-step explanation:

When the line is horizontal, the distance isn't changing, so she is stationary, so just add the total time the line is horizontal.

Likewise, all other lines represent traveling.

8 0

3 years ago

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Kevin Horn is the national sales manager for National Textbooks Inc. He has a sales staff of 40 who visit college professors all

Oduvanchick [21]

Answer:

Following are the solution to the given points:

Step-by-step explanation:

Given values:

$38, 40, 41, 45, 48, 48, 50, 50, 51, 51, 52, 52, 53, \\\\$ $54, 55, 55, 55, 56, 56, 57, 59, 59, 59, 62, 62, 62, \\ 63, 64, 65,66, 66, 67, 67, 69, 69, 71, 77, 78, 79, 79$

Total staff=40

In point a:

To calculate the median number first we arrange the value into ascending order and then collect the even numbers of calls that were also made. Its average of the middle terms is thus the median.

The midterms =55 and 59 so,

Median = $\frac{55+59}{2} = \frac{114}{2}=57$

In point b:

First-quarter $Q_1 = \frac{1}{4} \text{number of 8th call}$

$=\frac{1}{4} \times 30 th \\\\= 7.5 th$

The first quarterlies are $7.5th \ \ that \ is = (7+0.5)th\ \ term$

therefore the multiply of 0.5 by calculating the difference of the 7th and 8th term are:

$=0.5 \times 0= 0 \\\\\to Q_1 = 50+0=50 \\\\\to Q_3=\frac{3}{4} \text{number of 8th call} \\\\=\frac{3}{4} \times 30 th \\\\=22.5 \ th = (22+0.5)th \ \ term$

therefore the it is multiply by the 0.5 for the difference of the 22nd and 23rd term:

$= 0.5 \times 1=0.5 \\\\\to Q_3= 63+0.5=63.5$

In point c:

First decile $D_1 = \frac{1}{18} \text{number of 8th call}$

$= \frac{1}{10} \times 30 th \\\\= 3rd \ \ term\\\\\to D_1= 41 \\\\\to D_9= \frac{9}{10} \times 8\ th\ call\\\\=\frac{9}{10} \times 30 th \\\\=27th \ \ term\\\\\to D9 = 67$

In point d:

quartiles are:

$Q_1= 51.25 \\\\Q_3=66$

The right answers for the decile are:

$D_1=45.3 \\\\ D_9=76.4$

As for $D) P_{33} = 53.53$ will be available.

7 0

2 years ago

Can someone explain to me why are we adding 2kpi when we are doing zeros for sin and cos, but adding kpi when doing zeros for tg

makvit [3.9K]

on the first exercise, you got a solution angle of π/18, that's a good solution for the I Quadrant only, however, on a circle, we have angles that go from 0 to 2π, however we can always keep on going around and continute to 2π + π/2 or 3π or 4π, or 115π/3 or 1,000,000π/18 and so on, and we're really just going around the circle many times over, getting a larger and larger angle, same circular motion.

π/18 on that exercise works for the I Quadrant, however if we continue and go around say 2π, we'll find that 2π/3 + π/18 is a coterminal angle with π/18, and thus that angle has also the same sine value.

π/18 + 2kπ/3 , where k = integer, is a way to say, all angles around the circle that look like this have the same sine, namely

π/18 + 2(1)π/3

π/18 + 2(2)π/3

π/18 + 2(3)π/3

π/18 + 2(5)π/3

π/18 + 2(99999999)π/3

.....

so using the "k" as some sequence multiplier, is a generic notational way to say, "all these angles".

you'll also find that "n" is used as well for the same notation, say for example

2π/3 + 2πn.

8 0

3 years ago

Which answer is correct?<br> Thank you

stiv31 [10]

Answer:

Vertical angles

Step-by-step explanation:

: If two angles are vertical angles, then they're congruent (see the above figure). Vertical angles are one of the most frequently used things in proofs and other types of geometry problems, and they're one of the easiest things to spot in a diagram. Don't neglect to check for them!

4 0

2 years ago

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Here are two stories:

Dafna11 [192]

X is the ounces
Got this answer because you figured it out

5 0

2 years ago

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