how to find the middle term in the sequence 6, 30, 150, 750, …, 58, 593, 750

The polynomial approximate values are

a. The linear approximating polynomial is p1(x) = 2 - x

b.The quadratic approximating polynomial is  p2(x) = x² - 3*x + 3

c. The approximating polynomial value is p1(0.97) = 1.03; p2(0.97) = 1.0309

According to the statement

we have given that the f(x) = 1/x

And we have to find that polynomial approximate values which are written below

The linear approximating polynomial And quadratic approximating polynomial And approximate the given quantity of polynomials obtained in parts a. and b.

So, the given function is f(x) = 1/x

And

f'(x) =  -1/x²

f''(x) = 2/x³

a = 1

Now, we find the linear approximating polynomial

So,

a. The linear approximating polynomial is:

p1(x) = f(a) + f'(a)*(x - a)

p1(x) = 1/1 + -1/1² * (x - 1)

p1(x) = 1 - x + 1

p1(x) = 2 - x

And Now, we find the quadratic approximating polynomial

So,

b. The quadratic approximating polynomial is:

p2(x) = p1(x) + 1/2 * f''(a)*(x - a)²

p2(x) = 2 - x + 1/2 * 2/1³ * (x - 1)²

p2(x) = 2 - x + (x - 1)²

p2(x) = 2 - x + x² - 2*x + 1

p2(x) = x² - 3*x + 3

And Now, we find the approximating polynomial value

So,

c. approximate 1/0.97 using p1(x)

p1(0.97) = 2 - 0.97 = 1.03

approximate 1/0.97 using p2(x)

p2(0.97) = 0.97² - 3*0.97 + 3 = 1.0309

So, The polynomial approximate values are

a. The linear approximating polynomial is p1(x) = 2 - x

b.The quadratic approximating polynomial is  p2(x) = x² - 3*x + 3

c. The approximating polynomial value is p1(0.97) = 1.03; p2(0.97) = 1.0309.

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Question:

A. Find the linear approximating polynomial for the following function centered at the given point a. b. Find the quadratic approximating polynomial for the following function centeredat the given point a. c. Use the polynomials obtained in parts a. and b. to approximate the given quantity.

f(x)=1/x, a=1; approximate 1/0.97

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Answer:

w = 12

Step-by-step explanation:

We are given the equation:

[tex]\displaystyle{\dfrac{w-2}{4} = \dfrac{2w+1}{10}}[/tex]

First, clear out the denominators by multiplying both sides by LCM. In this scenario, our LCM is 40. Thus, multiply both sides by 40:

[tex]\displaystyle{\dfrac{w-2}{4} \cdot 40= \dfrac{2w+1}{10} \cdot 40}[/tex]

Simplify the expressions/equations:

[tex]\displaystyle{(w-2)\cdot 10= (2w+1)\cdot 4}\\\\\displaystyle{10w-20= 8w+4}[/tex]

Isolate w-variable:

[tex]\displaystyle{10w-20= 8w+4}\\\\\displaystyle{10w-8w=4+20}\\\\\displaystyle{2w=24}\\\\\displaystyle{w=12}[/tex]

Hence, the solution is w = 12

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he has 231 pennies..............

Answer:

A) y = 16/(x^2), B) 4/5

Step-by-step explanation:

For A, we can plug in some of the table values to check it. I will try 2 and 3

2. 4 = 16 / (2^2)

    16/4 = 4

3. 16/9 = 16 / (3^2)

   16/9 = 16/9

B) We can just input y into the formula 25 = 16 / (x^2)

This leaves us with +-4/5

Answer:

see explanation

Step-by-step explanation:

(a)

given y varies inversely as x² then the equation relating them is

y = [tex]\frac{k}{x^2}[/tex] ← k is the constant of variation

to find k substitute any ordered pair from the table into the equation

using (2, 4 ) , then

4 = [tex]\frac{k}{2^2}[/tex] = [tex]\frac{k}{4}[/tex] ( multiply both sides by 4 )

16 = k

y = [tex]\frac{16}{x^2}[/tex] ← equation of variation

(b)

when y = 25 , then

25 = [tex]\frac{16}{x^2}[/tex] ( multiply both sides by x² )

25x² = 16 ( divide both sides by 25 )

x² = [tex]\frac{16}{25}[/tex] ( take the square root of both sides )

x = ± [tex]\sqrt{\frac{16}{25} }[/tex] = ± [tex]\frac{4}{5}[/tex]

the positive value of x is x = [tex]\frac{4}{5}[/tex]

Let [tex]C[/tex] be the set of all students in the classroom.

Let [tex]P[/tex] and [tex]M[/tex] be the sets of students that pass physics and math, respectively.

We're given

[tex]n(C) = 60[/tex]

[tex]n(P \cap M') = 17[/tex]

[tex]n(P \cap M) = 25[/tex]

[tex]n((P \cup M)') = n(P' \cap M') = 9[/tex]

i. We can split up [tex]P[/tex] into subsets of students that pass both physics and math [tex](P\cap M)[/tex] and those that pass only physics [tex](P\cap M')[/tex]. These sets are disjoint, so

[tex]n(P) = n(P\cap M) + n(P\cap M') = 25 + 17 = \boxed{42}[/tex]

ii. 9 students fails both subjects, so we find

[tex]n(C) = n(P\cup M) + n(P\cup M)' \implies n(P\cup M) = 60 - 9 = 51[/tex]

By the inclusion/exclusion principle,

[tex]n(P\cup M) = n(P) + n(M) - n(P\cap M)[/tex]

Using the result from part (i), we have

[tex]n(M) = 51 - 42 + 25 = 34[/tex]

and so the probability of selecting a student from this set is

[tex]\mathrm{Pr}(M) = \dfrac{34}{60} = \boxed{\dfrac{17}{30}}[/tex]

Answer:

See below

Step-by-step explanation:

To pay the loan off more quickly.

To pay less overall interest.

Often get a lower interest rate for a shorter term loan.

The value of 'x' from the expression is 1/3

Given the expression;

[tex]x + \frac{x}{3} = \frac{4}{9}[/tex]

Find the LCM of the left side, we have

[tex]\frac{3x + x}{3} = \frac{4}{9}[/tex]

Cross multiply

[tex]9(4x) = 4 *3[/tex]

[tex]36x = 12[/tex]

Make 'x' the subject

[tex]x = \frac{12}{36}[/tex]

x = [tex]\frac{1}{3}[/tex]

Thus, the value of 'x' from the expression is 1/3

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Hello !

Answer:

[tex]\boxed{\sf x=\dfrac{1}{3} }[/tex]

Step-by-step explanation:

Our aim is to find the value of x that verifies the following equation :

[tex]\sf x+\frac{x}{3} =\frac{4}{9}[/tex]

Let's isolate x :

Multiply both sides by 3 :

[tex]\sf 3(x+\frac{x}{3} )=\frac{4}{9}\times 3\\ \sf 3x+x=\frac{4}{3}[/tex]

Now we can combine like terms :

[tex]\sf 4x=\frac{4}{3}[/tex]

Finally, let's divide both sides by 4 :

[tex]\sf \frac{4x}{4} =\frac{4}{3} \times \frac{1}{4} \\\boxed{\sf x=\dfrac{1}{3} }[/tex]

Have a nice day ;)

The percentage of tardiness among the 102 pupils is 2.94.

According to the query,

Course A: Total number of students = 102

                   Number of tardy students = 3

Percent of students tardy in course A = (number of tardy students/total  

                                                                   number of students) * 100

  =(3/102)*100

  = 2.941

  ≈2.94

In case of course B: Total number of students = 85 & tardy student = 5

So, the percent of tardy student = (5/85)*100 = 5.88 %

Therefore, it is concluded that the percentage of student tardiness in course A is 2.94.

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Answer: 1. A

2. A

Step-by-step explanation:

The cutoffs from the left and right sides of normal distribution at a 95-percent confidence level are 1.96.

The confidence level, which is used in statistics, describes the likelihood that the estimation of a statistical parameter's location in a sample survey is also true for the population.

Confidence levels must be decided upon in advance when surveying since they affect the survey's essential scope and error margin. Confidence intervals of 90, 95, and 99 percent are widely employed in surveys.

If the confidence level were set at 95%, there is a very good likelihood that the population's arithmetic mean, as a statistical number, will fall within the survey's established margins of error.

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The information to fill one the box regarding the proof include:

It should be noted that when two triangles of each corresponding sides are equal, then it's said that they are similar.

Here AB = CD, and AD = CB as they illustrate the fact that they are parallel.

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The error in the measurement of its volume is 3%.

The error in the measurement of its volume is:​

Volume[tex]=\frac{4\pi r^{3} }{3}[/tex]

Δv/v*100=3 ΔR/R*100

[tex]3*\frac{1}{100} *100[/tex]

=3

The error in the measurement of its volume is 3%.

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Simplifying the expression gives 36x^2 - 82x - 10 = 0

Given the expression;

4x-5/3+2x=7+2/9x+2

[tex]\frac{4x - 5}{3 + 2x} = \frac{9}{9x + 2}[/tex]

Cross multiply

[tex]4x - 5( 9x + 2) = 9 (3 + 2x)[/tex]

Expand the bracket

[tex]36x^2 + 8x - 45x - 10 = 27x + 18x[/tex]

Collect like terms

36x^2 + 8x - 45x - 27x - 18x = 0

Add the like terms

36x^2- 82x - 10 = 0

Thus, simplifying the expression gives 36x^2- 82x - 10 = 0

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Prediction of the value of the dependent variable outside the experimental region is called extrapolation.

According to the question,

Prediction of the value of the dependent variable outside the experimental region is called extrapolation.

Extrapolation is the statistical method beamed at understanding the unknown data from the known data.

Hence, prediction of the value of the dependent variable outside the experimental region is called extrapolation.

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Answer:

there should have been names of every line and joining parts

Answer:

12

Step-by-step explanation:

The area of a triangle is calculated with the following formula:

[tex]\frac{1}{2} *b*h[/tex] (b: base, h: height (or legs))

We can use this formula to find the length of the other leg:

30 = [tex]\frac{1}{2} *b*h[/tex] multiply both sides with 2 to get rid of fraction

60 = b*h one of the leg's length is given as 5 so

60 = 5*h divide both sides by 5

12 = h is the length of the missing leg.

The measure of angle G in terms of x is x+x degrees

The measure of angle F and angle D is 90 degrees so that;

<GFD = <GDF = 90 - x

Since the sum of angle in a triangle is 180 degrees, hence;

<G + 90 - x + 90 - x = 180

<G + 180 - 2x = 180

<G = 2x

<G = x + x

Hence the measure of angle G in terms of x is x+x degrees

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Answer:A

Step-by-step explanation: math and me go together like mustard and pees

Answer:

range of the given set is { 7,2,4,2,1}

Step-by-step explanation:

. Hello !

When the above kind of set is given the second values of each elements belongs to the range or the y whereas, the first values of the elements belongs to the domain or x.

The total area of the square based prism is 413.7 square units

The given parameters are:

Base length, a = 10

Height, h = 11

The total surface area is calculated as:

[tex]A = a^2 + 2a\sqrt{\frac{a^2}{4} + h^2[/tex]

This gives

[tex]A = 10^2 + 2 *10\sqrt{\frac{10^2}{4}+11^2[/tex]

So, we have:

[tex]A = 100 + 20\sqrt{246[/tex]

Evaluate

A = 413.7

Hence, the total area of the square based prism is 413.7 square units

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Answer:

Cookies = 44

Step-by-step explanation:

universal U=60

let the number of cookies be C,

n(C)=(3*b)-4

and the number of brownies be b,

n(b)=?

n(buc)=60

b+c=60

then, c=60-b

substitute for c=60-b.

60-b=3b-4

60+4=3b+b

64=4b

b=16

c=60-b

c= 60-16

c=44

68% of the daily phone calls answered by the company are between 47 and 53.

Empirical rule states that for a normal distribution, 68% of the data are within one standard deviation from the mean, 95% of the data are within two standard deviation from the mean and 99.7% of the data are within three standard deviation from the mean.

Given mean of 50 and a standard deviation of 3

68% are within one standard deviation from mean = mean ± standard deviation = 50 ± 3 = (47, 53)

68% of the daily phone calls answered by the company are between 47 and 53.

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Answer:

B,C and E

Step-by-step explanation:

The period can be calculated by dividing 2pi by the coefficfient of the x, so in this case 1. And we'll get 2pi/1 = 2pi.

The function clearly has a minimum of -2, you can see that from the table.

Same thing for the maximum = 4.

Answer:

See Photo

Step-by-step explanation:

Plato/Edmentum

Given the average speed and distance covered, the time taken for the driver to complete the road trip is 5.7 hours

Speed is simply referred to as distance traveled per unit time.

Mathematically, Speed = Distance ÷ time.

Given the data in the question;

We substitute into our equation above.

Speed = Distance ÷ time

70 miles per hour = 400 mile ÷ Elapsed time

Elapsed time = 400 miles ÷ 70 miles per hour

Elapsed time = 5.7 hours

Given the average speed and distance covered, the time taken for the driver to complete the road trip is 5.7 hours.

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The answers have been shown below.

Questions regarding the time spent by Jack jogging and bike riding are needed to be answered.

(A) The equations are [tex]x+y=75, y=15+x[/tex]

Time spent jogging is 30 minutes

The total time would be [tex]45+60=105[/tex] minutes which is not equal to 75 minutes.

(B) Let [tex]x[/tex] be the time spent jogging.

[tex]y[/tex] be the time spent bike riding.

[tex]x+y=75\\y=15+x\\x+15+x=75\\2x+15=75\\x=\frac{75-15}{2} =30[/tex]

Time spent jogging is 30 minutes.

[tex]y=60\\x+y=75[/tex]

(C) If he rides his bike 15 minutes longer than he jogs then he would have to jog [tex]60-15 = 45[/tex] minutes.

Therefore, the total time would be [tex]45+60=105[/tex] minutes which is not equal to 75 minutes.

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When everyone is sitting down, the number of free seats is 26, using arithmetic addition and subtraction.

The two main arithmetic operations that we learn to add and subtract two or more integers or other mathematical values are arithmetic addition and subtraction. The addition symbol is the plus sign (+), and the subtraction symbol is the minus sign (-). (minus sign).

The initial number of passengers on board=29

After 4 people get off at the first stop,

Using arithmetic addition and subtraction, we get,

Number of passengers left=29-4=25

After 11 people get on,

Number of passengers=25+11=36

At the second stop, when 7 people get off,

Number of passengers=36-7=29

When 23 people get on,

Number of passengers=29+23=52

Total number of passenger seats in the bus=78

Number of free seats, when everyone is sitting=78-52

                                                                              =26

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Answer:

I may have misubderstood the question, so review carefully.

width = 380 m

length is 420 m

Step-by-step explanation:

Let w and l stand for the width and length of the stadium.

We are told that w = l - 40 m

We also learn that l = 420 m

w = l - 40 m

w = (420 m) - 40 m

width = 380 m

length is 420 m

Similarities and differences between solving an absolute value equation and solving an absolute value inequality are shown below.

Similarities and differences between solving an absolute value equation and solving an absolute value inequality:

1) The absolute value equation of a number is simply the number's distance from zero.

2) Absolute value inequalities and linear inequalities share the fact that they both have two variables and so require a second equation to obtain the variables.

Therefore, the similarities and differences between solving an absolute value equation and solving an absolute value inequality are shown.

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Step-by-step explanation:

1. Not neccesarily true

2. Not necessarily true

3. True because angles in a triangle add to 180°

4. Not necessarily true

5. True because the medians are being constructed, and the three medians of a triangle are always concurrent.

6. Not necessarily true

The statements that are true about the triangle are:

Option C: m∠P + m∠Q + m∠R = 180°

Option E: PT, QU, and RS intersect at the same point.

1. m∠Q = m∠R: This is not true because there is no indication that the angles are equal.

2.The length of QU is half the length of RP: This is not true because there is no length given to show that measurement.

3. m∠P + m∠Q + m∠R = 180°:

This is true because the sum of angles in a triangle add to 180°

4. QU ≅ RS:

This is not true because we are not told that they are congruent

5. PT, QU, and RS intersect at the same point:

This is true because the medians are being constructed, and the three medians of a triangle are always concurrent.

6. The sum of the lengths of QU and RS is equal to the length of PT: This is not true because we are not told that.

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The mean of the data set is 81.6.

The median of the data set is 80.

The mean is the average of a data set. Therefore,

Mean = (70 x 2 + 75 x 8 + 80 x 6 + 85 x 3 + 90 x 9) / 28 = 81.6.

Hence,

mean = 81.6

Median is a statistical measure that determines the middle value of a dataset listed in ascending order .

Therefore,

28 / 2 = 14

median = 80 + 80 / 2 = 160 / 2 = 80

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Answer:

-31, /~26, -5, 6

Step-by-step explanation: