All categories

2 years ago

Which is the polynomial function of lowest degree with rational real coefficients, a leading coefficient of 3 and roots 5 and 2?

2 answers:

6 0

A root of a polynomial doesn't mean it is a square root of!!!!!!

P(x)=3(x-5)(x-2)=3x²-21x+30

f (x)= 3 x power point 2 - 21 x + 30

Alex Ar [27]2 years ago

4 0

Answer:

the polynomial is: $3x^2-21x+30$

Step-by-step explanation:

we know that the polynomial function p(x) of lowest degree with roots as 'a' and 'b' and leading coefficient as 'c' is given by: $p(x)=c(x-a)(x-b)$

here we are given that the roots are 5 and 2 and the leading coefficient is 3.

so the polynomial p(x) of lowest degree with the above properties is: $p(x)=3(x-5)(x-2)$

$p(x)=3(x^2-5x-2x+10)$

$p(x)=3(x^2-7x+10)$

$p(x)=3x^2-21x+30.$ .

You might be interested in

You have a function that takes in an X value and produces a Y value. The Y value equals 24 times the X value, plus 4?More. What'

guapka [62]

Answer:

100

Step-by-step explanation:

The function takes in an X value and produces a Y value.

The Y value equals 24 times the X value plus 4 more.

This means that:

Y = 24X + 4

When the X value equals 4, the Y value will be:

Y = 24(4) + 4

Y = 96 + 4

Y = 100

When the X value is 4, the Y value is 100.

7 1

1 year ago

Read 2 more answers

Find the length of side AB Give answer to 3 significant figures

Rina8888 [55]

Answer:

AB = 8.857 cm

Step-by-step explanation:

Here, we are given a right angle $\triangle ABC$ in which we have the following things:

$\angle A = 90 ^\circ\\\angle C = 41 ^\circ\\\text{Side }BC = 13.5 cm$

Side BC is the hypotenuse here.

We have to find the side AB.

Trigonometric functions can be helpful to find the value of Side AB here.

Calculating $\angle B$ :

Sum of all the angles in $\triangle ABC$ is $180^\circ$ .

$\Rightarrow \angle A + \angle B + \angle C = 180^\circ\\\Rightarrow 90^\circ + \angle B + 41^\circ = 180^\circ\\\Rightarrow \angle B = 49^\circ$

We know that cosine of an angle is:

$cos \theta = \dfrac{\text{Base}}{\text{Hypotenuse}}\\\Rightarrow cos B = \dfrac{AB}{BC}\\\Rightarrow cos 49^\circ = \dfrac{AB}{13.5}\\\Rightarrow AB = 13.5 \times 0.656\\\Rightarrow AB = 8.857 cm$

So, side AB = 8.857 cm .

6 0

1 year 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

Peter decided to buy a new car. He made a $2,160 down payment and then took a 48-month loan. The total amount for the car, plus

slava [35]

16560-2160=14400, 14400=48x, x=$300

3 0

2 years ago

Read 2 more answers

lev has 9 tens. jim has 1 ten. they put all their tens together. dawn puts 9 ones with the tens. what number did they make?

MatroZZZ [7]

The answer to this 110

3 0

2 years ago