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2 years ago

What are the solutions of the equation x4 + 3x2 + 2 = 0? Use u substitution to solve.

Mathematics

2 answers:

myrzilka [38]2 years ago

4 0

I hope this helps you

babymother [125]2 years ago

3 0

We can use t=x^2 to solve this
Once we do that we will have simple square equation which we know how to solve.

t^2 + 3t + 2 = 0

t1 = -1
t2 = -2

x1 = √-1 = i
x2 = -i

x3 = √-2 = i√2
x4 = -i√2

Make sure you know that i^2 = -1 and (-i)^2 = -1 which gives us solutions we got...

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On a coordinate plane, 5 trapezoids are shown. Trapezoid L M N P is in quadrant 2 and has points (negative 5, 1), (negative 5, 4

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Answer:

The answer is figure C

Step-by-step explanation:

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2 years ago

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Huang buys 3 shirts that each cost the same amount, a pair of pants that cost $12, and he pays with a $100 bill. Which

nekit [7.7K]

Answer:

(100 – 12) – 3x

Step-by-step explanation:

(100 – 12) – 3x would be the correct answer because:

100 is the total amount he paid

12 is the cost of pants

because we don't know the price of shirts yet so it would be called x

He buy in total of 3 shirts so the equation for the price would be 3 * x = 3x

So it would be:

total money paid - pants costs - shirts cost

= 100 - 12 - 3x or (100 - 12) -3x

Hope this helped :3

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2 years ago

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Thirty percent of the students in a high school face a disciplinary action of some kind before they graduate. Of those students,

MAXImum [283]

Answer:

D. 54%

Step-by-step explanation:

For simplicity, lets assume we have 100 students in high school.

30% of students face disciplinary action, therefore;

⇒ (30/100) x 100 = 30 students.

of these 30 students, 40% go to college. therefore;

⇒ (40/100) x 30 = 12 students (go to college)

Also 70% of students do not face disciplinary action, therefore;

⇒ (70/100) x 100 = 70 students.

of these 70 students. 60% go to college. therefore;

⇒ (60/100) x 70 = 42 students (go to college)

Total number of students who go to college

⇒ 12 + 42 = 54 students

% of students who go to college

⇒ 54/100 = 54%

7 0

2 years ago

According to a Los Angeles Times study of more than 1 million medical dispatches from 2007 to 2012, the 911 response time for me

Reil [10]

Answer:

a) $\bar X=10.65$

$Median =\frac{10.7+10.7}{2}=10.7$

$Mode= 10.7$

b) $Range = 11.8-8.3=3.5$

$s= 0.948$

c) $IQR = Q_3 -Q_1 = 11.05-10.55=0.5$

And we can find the usual limits with:

$Lower = Q_1 -1.5 IQR = 10.55 -1.5*0.5=9.8$

$Upperer = Q_3 +1.5 IQR = 10.55 +1.5*0.5=11.3$

And since 8.3 <9.8 we can consider this value too low or as an outlier for this case.

d) The mean for this case was 10.65 and the usual values are between 9.8 and 11.3, so as we can see all are above the specified value of 6 minutes, and we can conclude that the times are not satisfying the quality standards for this case.

And they should be considered apply some strategies to reduce the response time, adding more stations around points selected at the city could be useful in order to reduce the response time.

Step-by-step explanation:

We have the following data:

11.8 10.3 10.7 10.6 11.5 8.3 10.5 10.9 10.7 11.2

Part a

We can calculate the sample mean with the following formula:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

And if we replace we got: $\bar X=10.65$

For the median we need to sort the values on increasing order and we have:

8.3 10.3 10.5 10.6 10.7 10.7 10.9 11.2 11.5 11.8

Since n =10 we can calculate the median as the average between the 5th and 6th position of the dataset ordered and we got:

$Median =\frac{10.7+10.7}{2}=10.7$

The mode would be the most repeated value on this case:

$Mode= 10.7$

Part b

The range is defined as $Range =Max-Min$ and if we replace we got:

$Range = 11.8-8.3=3.5$

We can calculate the standard deviation with the following formula:

$s= sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

And if we replace we got:

$s= 0.948$

Part c

For this case we can use the IQR method in order to determine if 8.3 is an outlier or not.

We can calculate the first quartile with these values: 8.3 10.3 10.5 10.6 10.7 10.7 and $Q_1= \frac{10.6+10.7}{2}=10.55$

And for the Q3 we can use: 10.7 10.7 10.9 11.2 11.5 11.8 and we got $Q_3 = \frac{10.9+11.2}{2}=11.05$

Then we can find the IQR like this:

$IQR = Q_3 -Q_1 = 11.05-10.55=0.5$

And we can find the usual limits with:

$Lower = Q_1 -1.5 IQR = 10.55 -1.5*0.5=9.8$

$Upperer = Q_3 +1.5 IQR = 10.55 +1.5*0.5=11.3$

And since 8.3 <9.8 we can consider this value too low or as an outlier for this case.

Part d

The mean for this case was 10.65 and the usual values are between 9.8 and 11.3, so as we can see all are above the specified value of 6 minutes, and we can conclude that the times are not satisfying the quality standards for this case.

And they should be considered apply some strategies to reduce the response time, adding more stations around points selected at the city could be useful in order to reduce the response time.

6 0

2 years ago

A square lawn has area 800ft^2. A sprinkler placed at the center of the lawn sprays water in a circular pattern as shown in the

mestny [16]

I'm pretty sure your radius is 14.14'

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2 years ago