The black graph is the graph of y =f(x). Choose the equation for the red graph. A. y - 1 = f() B. y + 1 = f() C. = f(x - 1) D. f(x)​

The zeros of the given equation are -5  and -7

Quadratic equations are equations that has a leading degree of 2. Given the factors of a quadratic equation as expressed below;

(-3x - 15)(x+7) = 0

The expressions -3x -15 and x + 7 are the factors of the equation. Equating both factors to zero

-3x - 15 = 0

Add 15 to both sides of the equation

-3x -15 + 15 = 0 + 15

-3x = 15

Divide both sides of the equation by -3

-3x/-3 = 15/-3

x = -5

Similarly;

x + 7 = 0

x = -7

Hence the zeros of the given equation are -5  and -7

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The quotient in lowest term exists 2/3.

An improper fraction exists as a fraction whose numerator exists equivalent to, larger than, or of equivalent or higher degree than the denominator.

Given the quotient: [tex]$4\frac{2}{3} \div 7[/tex]

Write [tex]$\frac{2}{3}[/tex] in improper fraction

[tex]$4\frac{2}{3}= \frac{14}{3}[/tex]

Dividing throughout by 7 on both sides of the equation, we get

[tex]$4\frac{2}{3} \div 7 = \frac{14}{3}\div 7[/tex]

Change the division sign to multiplication by taking the reciprocal of 7

[tex]$\frac{14}{3}\div 7 = \frac{14}{3} \times \frac{1}{7}[/tex]

Simplifying the above equation, we get

[tex]$\frac{14}{3}\div 7 = \frac{2}{3}[/tex]

Therefore, the quotient in lowest term exists 2/3.

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Answer:

[tex]\dfrac{3}{2} \ln |x-4| - \dfrac{1}{2} \ln |x+2| + \text{C}[/tex]

Step-by-step explanation:

Fundamental Theorem of Calculus

[tex]\displaystyle \int \text{f}(x)\:\text{d}x=\text{F}(x)+\text{C} \iff \text{f}(x)=\dfrac{\text{d}}{\text{d}x}(\text{F}(x))[/tex]

If differentiating takes you from one function to another, then integrating the second function will take you back to the first with a constant of integration.

Given indefinite integral:

[tex]\displaystyle \int \dfrac{x+5}{(x-4)(x+2)}\:\:\text{d}x[/tex]

Take partial fractions of the given fraction by writing out the fraction as an identity:

[tex]\begin{aligned}\dfrac{x+5}{(x-4)(x+2)} & \equiv \dfrac{A}{x-4}+\dfrac{B}{x+2}\\\\\implies \dfrac{x+5}{(x-4)(x+2)} & \equiv \dfrac{A(x+2)}{(x-4)(x+2)}+\dfrac{B(x-4)}{(x-4)(x+2)}\\\\\implies x+5 & \equiv A(x+2)+B(x-4)\end{aligned}[/tex]

Calculate the values of A and B using substitution:

[tex]\textsf{when }x=4 \implies 9 = A(6)+B(0) \implies A=\dfrac{3}{2}[/tex]

[tex]\textsf{when }x=-2 \implies 3 = A(0)+B(-6) \implies B=-\dfrac{1}{2}[/tex]

Substitute the found values of A and B:

[tex]\displaystyle \int \dfrac{x+5}{(x-4)(x+2)}\:\:\text{d}x = \int \dfrac{3}{2(x-4)}-\dfrac{1}{2(x+2)}\:\:\text{d}x[/tex]

[tex]\boxed{\begin{minipage}{5 cm}\underline{Terms multiplied by constants}\\\\$\displaystyle \int ax^n\:\text{d}x=a \int x^n \:\text{d}x$\end{minipage}}[/tex]

If the terms are multiplied by constants, take them outside the integral:

[tex]\implies \displaystyle \dfrac{3}{2} \int \dfrac{1}{x-4}- \dfrac{1}{2} \int \dfrac{1}{x+2}\:\:\text{d}x[/tex]

[tex]\boxed{\begin{minipage}{5 cm}\underline{Integrating}\\\\$\displaystyle \int \dfrac{f'(x)}{f(x)}\:\text{d}x=\ln |f(x)| \:\:(+\text{C})$\end{minipage}}[/tex]

[tex]\implies \dfrac{3}{2} \ln |x-4| - \dfrac{1}{2} \ln |x+2| + \text{C}[/tex]

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For an alternative approach, expand and complete the square in the denominator to write

[tex](x-4)(x+2) = x^2 - 2x - 8 = (x - 1)^2 - 9[/tex]

In the integral, substitute [tex]x - 1 = 3 \sin(u)[/tex] and [tex]dx=3\cos(u)\,du[/tex] to transform it to

[tex]\displaystyle \int \frac{x+5}{(x - 1)^2 - 9} \, dx = \int \frac{3\sin(u) + 6}{9 \sin^2(u) - 9} 3\cos(u) \, du \\\\ ~~~~~~~~~~~~ = - \int \frac{\sin(u) + 2}{\cos(u)} \, du \\\\ ~~~~~~~~~~~~ = - \int (\tan(u) + 2 \sec(u)) \, du[/tex]

Using the known antiderivatives

[tex]\displaystyle \int \tan(x) \, dx = - \ln|\cos(x)| + C[/tex]

[tex]\displaystyle \int \sec(x) \, dx = \ln|\sec(x) + \tan(x)| + C[/tex]

we get

[tex]\displaystyle \int \frac{x+5}{(x - 1)^2 - 9} \, dx = \ln|\cos(u)| - 2 \ln|\sec(u) + \tan(u)| + C \\\\ ~~~~~~~~~~~~ = - \ln\left|\frac{(\sec(u) + \tan(u))^2}{\cos(u)}\right|[/tex]

Now, for [tex]n\in\Bbb Z[/tex],

[tex]\sin(u) = \dfrac{x-1}3 \implies u = \sin^{-1}\left(\dfrac{x-1}3\right) + 2n\pi[/tex]

so that

[tex]\cos(u) = \sqrt{1 - \dfrac{(x-1)^2}9} = \dfrac{\sqrt{-(x-4)(x+2)}}3 \implies \sec(u) = \dfrac3{\sqrt{-(x-4)(x+2)}}[/tex]

and

[tex]\tan(u) = \dfrac{\sin(u)}{\cos(u)} = -\dfrac{x-1}{\sqrt{-(x-4)(x+2)}}[/tex]

Then the antiderivative we found is equivalent to

[tex]\displaystyle - \int \frac{x+5}{(x - 1)^2 - 9} \, dx = - \ln\left|-\frac{3(x+2)}{(x-4) \sqrt{-(x-4)(x+2)}}\right| + C[/tex]

and can be expanded as

[tex]\displaystyle - \int \frac{x+5}{(x - 1)^2 - 9} \, dx = -\ln\left| \frac{3(x+2)^{1/2}}{(x-4)^{3/2}}\right| + C \\\\ ~~~~~~~~~~~~ = - \ln\left|(x+2)^{1/2}\right| + \ln\left|(x-4)^{3/2}\right| + C \\\\ ~~~~~~~~~~~~ = \boxed{\frac32 \ln|x-4| - \frac12 \ln|x+2| + C}[/tex]

Answer: (2,9)

Step-by-step explanation:

The point lies in the region that is shaded by both inequalities.

The inequality in the box has to be written as

x² + 2x - 80 ≤ - 65

We have

(10 + x)1 * (16-2x) ≥ 130

Next we would have to open the bracket

160 + 16x - 20x - 2x² ≥ 130

Then we would have to arrange the equation

- 2x² - 4x + 160 ≥ 130

Divide the equation by two

- x² - 2x + 80 ≥ 65

This is arranged as

x² + 2x - 80 ≤ - 65

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Answer:

Option D

Step-by-step explanation:

Using the surd law :

[tex]\sqrt{ab} = \sqrt{a}\sqrt{b}[/tex]

We can find the largest square number that goes into 75 :

Let's write the multiples of 75 :

1 , 75

3 , 25

5 , 15

The only square number is 25

So using the law mentioned above we split √75 into :

√25√3

The square root of 25 is 5

Now we have our final answer of 5√3

Hope this helped and have a good day

The simplified form of expression √75 is 5√3.

Option D is the correct answer.

We have,

To simplify √75, we can factor it into its prime factors and then take the square root:

√75 = √(3 * 5 * 5)

= √(3 x 5²)

Take out the perfect square factor from under the square root:

= √3 x √5²

= √3 x 5

= 5√3

Thus,

The simplified form of expression √75 is 5√3 which is option D.

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Answer:

[tex]\displaystyle x=\frac{5}{4},\;\;1\frac{1}{4}, \;\; or \;\; 1.25[/tex]

Step-by-step explanation:

    To solve for x, we need to isolate the x variable.

    Given:

[tex]\displaystyle x+\frac{1}{2} =\frac{7}{4}[/tex]

    Subtract [tex]\frac{1}{2}[/tex] from both sides of the equation:

[tex]\displaystyle (x+\frac{1}{2})-\frac{1}{2} =(\frac{7}{4})-\frac{1}{2}[/tex]

[tex]\displaystyle x=\frac{7}{4}-\frac{1}{2}[/tex]

    Now, we will create common denominators to simplify.

[tex]\displaystyle x=\frac{7}{4}-\frac{2}{4}[/tex]

[tex]\displaystyle x=\frac{5}{4}[/tex]

A polynomial function exists a function that contains only non-negative integer powers or only positive integer exponents of a variable in an equation like the quadratic equation, cubic equation, etc.

The variety of function kinds can model circumstances in the real world. Properties or key components of functions create more or less suitable models, relating to the circumstances.

Linear: These relationships maintain constant rates of change. The distance traveled by a car moving at a constant speed over time exists linear.

Exponential: Connections that show a percent change exist exponential. The growth of a financial investment accumulating interest over time exists exponential.

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Using the Laplace transform, the value of y' + y = (t − 1), y(0) = 5 is y(t) = 5e ^ -t + u (t - 1)e^(1-t)

Laplace rework is an critical rework approach that is in particular useful in fixing linear normal equations. It unearths very huge applications in  regions of physics, electrical engineering, control optics, arithmetic and sign processing.

y' + y = (t − 1)

y (0) = 5

Taking the Laplace transformation of the differential equation

⇒sY(s) - y (0) + Y(s) = e-s

⇒(s + 1)Y(s) = (5+ e^-s)/s + 1

⇒y(t) = L^-1{5/s+1} + {e ^-s/s + 1}

⇒y(t) = 5 e^-t + u(t -1)e^1-t

The Laplace remodel method, the feature within the time area is transformed to a Laplace characteristic within the frequency domain. This Laplace feature will be inside the shape of an algebraic equation and it can be solved easily.

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The dipping sauce which has a stronger mustard flavor between burger barn and be sandwich town is burger barn

Burger bun:

Mustard : honey

= 2 : 4

= 2/4

= 1/2

= 0.5

Sandwich:

Mustard : honey

= 2 : 8

= 2/8

= 1/4

= 0.25

Therefore, burger barn has a more stronger mustard flavor of dipping sauce between burger barn and be sandwich town.

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Answer:

Step-by-step explanation:

The two dipping sauce have same taste.

Answer:

see explanation

Step-by-step explanation:

total parts = 3 + 18 + 6 + 3 = 30

3 grams of silver = [tex]\frac{3}{30}[/tex] = [tex]\frac{1}{10}[/tex]

18 grams of copper = [tex]\frac{18}{30}[/tex] = [tex]\frac{3}{5}[/tex]

6 grams of aluminium = [tex]\frac{6}{30}[/tex] = [tex]\frac{1}{5}[/tex]

3 grams of zinc = [tex]\frac{3}{30}[/tex] = [tex]\frac{1}{10}[/tex]

Step-by-step explanation:

using gp formula

tn=ar^n-1

which our a which is the first term is = -3

our r which is t2/t1=-6

t8=(-3)(-6)^8-1

=(-3)(-6)^7

=(-3)(-279936)

=839808

Split up the interval [0, 3] into 3 equally spaced subintervals of length [tex]\Delta x = \frac{3-0}3 = 1[/tex]. So we have the partition

[0, 1] U [1, 2] U [2, 3]

The left endpoint of the [tex]i[/tex]-th subinterval is

[tex]\ell_i = i - 1[/tex]

where [tex]i\in\{1,2,3\}[/tex].

Then the area is given by the definite integral and approximated by the left-hand Riemann sum

[tex]\displaystyle \int_0^3 f(x) \, dx \approx \sum_{i=1}^3 f(\ell_i) \Delta x \\\\ ~~~~~~~~~~ = \sum_{i=1}^3 (i-1)^3 \\\\ ~~~~~~~~~~ = \sum_{i=0}^2 i^3 \\\\ ~~~~~~~~~~ = 0^3 + 1^3 + 2^3 = \boxed{9}[/tex]

The x value or observed value corresponding to  z-score, z = 1.50 is 130.

According to the question.

For a population with µ = 100 and σ = 20.

Since,  we know that

The z-score is a statistical evaluation of a value's correlation to the mean of a collection of values, expressed in terms of standard deviation.

And it is given by

z = (x - μ) / σ

Where,

x is the observed value.

μ is the mean.

and, σ is the standard deviation.

Therefore, the x value or observed value corresponding to z = 1.50 is given by

[tex]1.50 = \frac{x -100}{20}[/tex]

⇒ 1.50 × 20 = x - 100

⇒ 30 = x - 100

⇒ x = 30 + 100

⇒ x = 130

Hence, the x value or observed value corresponding to  z-score, z = 1.50 is 130.

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Answer:

mass of a clay pot = 4.95 kg

Kindly award branliest

Step-by-step explanation:

Let the mass of a clay pot be x

Let the mass of a metal pot be y

Thus; 2x + 2y = 13.2

And ;

x = 3 times y

x = 3y

2x + 2y = 13.2

2(3y) + 2y = 13.2

6y + 2y = 13.2

8y = 13.2

y = 13.2/8 = 1.65

x = 3y = 3(1.65) = 4.95

mass of a clay pot = 4.95 kg

The box and whisker plot is plotted with minimum value 24, maximum value 58, first quartile 29, third quartile 52 and median 40.

Given that, the ages of 15 music teachers in a school district are 24, 26, 27, 29, 29, 32, 37, 40, 40, 41, 45, 52, 56, 56, 58.

A box and whisker plot also called a box plot displays the five-number summary of a set of data. The five-number summary is the minimum, first quartile, median, third quartile, and maximum. In a box plot, we draw a box from the first quartile to the third quartile. A vertical line goes through the box at the median.

Minimum value = 24

Maximum value = 58

First quartile = 29

Third quartile = 52

Median = 40

Therefore, the box and whisker plot is plotted with minimum value 24, maximum value 58, first quartile 29, third quartile 52 and median 40.

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The answer is B.

Since ΔQRS ~ ΔXYZ, the value of tan(Q) is :

The z-score corresponding to a sample mean of m = 69 is -0.167

In this problem, we have been given :

population mean (μ) = 70, standard deviation (σ) = 12,  sample size (n) = 4, sample mean (m) = 69

We know that, the Z-score measures how many standard deviations the measure is from the mean.

Also, the formula when calculating the z-score of a sample with known population standard deviation is:

[tex]Z=\frac{m-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex]

where z = standard score

μ = population mean

σ = population standard deviation

m = the sample mean

and [tex]\frac{\sigma}{\sqrt{n} }[/tex] is the Standard Error of the Mean for a Population

First we find the Standard Error of the Mean for a Population

σ /√n

= 12 / √4

= 12 / 2

= 6

So, the z-score would be,

⇒ [tex]Z=\frac{m-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex]

⇒ [tex]Z=\frac{69-70}{6 }[/tex]

⇒ Z = -1/6

⇒ Z = -0.167

Therefore, the z-score corresponding to a sample mean of m = 69 is -0.167

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Answer:

y=-\frac{2}{5}x+\frac{19}{5}y=−52x+519

Further explanation:

We have to find the equation of the line first to graph the line.

The general form of slope-intercept form of equation of line is:

y=mx+by=mx+b

Given

m=-\frac{2}{5}m=−52

Putting the value of slope in the equation

y=-\frac{2}{5}x+by=−52x+b

To find the value of b, putting the point (-3,5) in equation

\begin{gathered}5=-\frac{2}{5}(-3)+b\\5=\frac{6}{5}+b\\5-\frac{6}{5}+b\\b=\frac{25-6}{5}\\b=\frac{19}{5}\end{gathered}5=−52(−3)+b5=56+b5−56+bb=525−6b=519

Putting the values of b and m

y=-\frac{2}{5}x+\frac{19}{5}y=−52x+519

The average median and mode price is $18. Option C is correct.

Given that certain item is available at 7 stores, three stores sell them for $20, two stores sell them for $15, one store sells them for $13, and one sells them for $16.

The mean is the average of the given numbers and is calculated by dividing the sum of the given numbers by the total number of numbers.

Firstly, we will find the median of the given items by arranging the given numbers in ascending order, we get

13,15,15,16,20,20,20

To find the median use the formula (n+1)/2, where n is the number of values in your dataset.

(7+1)/2=8/2=4

In the ascending order numbers 4th term is 16.

So, median is 16

Mode is the highest repeating term in the set or numbers.

So, here mode is 20

Now, we will calculate the average of median and mode, we get

Average=(median +mode)/2

Average=(16+20)/2

Average=18

Hence, the average (arithmetic mean) of the median price and the mode price where certain item is available at 7 stores, three stores sell it for $20, two stores sell it for $15, one store sells it for $13, and one sells it for $16 is $18.

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Answer:

Option 4

Step-by-step explanation:

By the Pythagorean identity, and the fact we are in the first quadrant,

[tex]\cos \theta=\frac{4}{5}[/tex]

Using the double angle formula for cosine,

[tex]\cos 2\theta=2\cos^{2} \theta-1=2\left(\frac{4}{5} \right)^2 -1=\frac{7}{25}[/tex]

Answer:

Step-by-step explanation:

Givens

Time 1 = 11.74

Time 2 = 11.26

Time 3 = 12.12           Add

Solution

11.74 + 11.26 + 12.12 =

Total Time = 35.12

35.12 miles is the total number of minutes that it took krissy to run these three miles.

What is a simple definition of time?

The measured or measurable period during which an action, process, or condition exists or continues : duration. b : a nonspatial continuum that is measured in terms of events which succeed one another from past through present to future.

Krissy ran three miles one morning she ran the first mile in time 1 = 11.74

Krissy ran three miles one morning she ran the first mile in time 2 = 11.26

Krissy ran three miles one morning she ran the first mile in time 3 = 12.12          

To get total time , we have to add all the time 1,2,3

Total time = Time1 + Time2 + Time3

                  = 11.74 + 11.26 + 12.12

Total Time =  35.12 miles

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Answer:

[tex]\displaystyle{\sin \theta = \dfrac{2ab}{a^2+b^2}}\\\\\displaystyle{\cos \theta = \dfrac{a^2-b^2}{a^2+b^2}}\\\\\displaystyle{\csc \theta = \dfrac{a^2+b^2}{2ab}}\\\\\displaystyle{\sec \theta = \dfrac{a^2+b^2}{a^2-b^2}}\\\\\displaystyle{\cot \theta = \dfrac{a^2-b^2}{2ab}}[/tex]

Step-by-step explanation:

We are given that:

[tex]\displaystyle{\tan \theta = \dfrac{2ab}{a^2-b^2}}[/tex]

To find other trigonometric ratios, first, we have to know that there are total 6 trigonometric ratios:

[tex]\displaystyle{\sin \theta = \sf \dfrac{opposite}{hypotenuse} = \dfrac{y}{r}}\\\\\displaystyle{\cos \theta = \sf \dfrac{adjacent}{hypotenuse} = \dfrac{x}{r}}\\\\\displaystyle{\tan \theta = \sf \dfrac{opposite}{adjacent} = \dfrac{y}{x}}\\\\\displaystyle{\csc \theta = \sf \dfrac{hypotenuse}{opposite} = \dfrac{r}{y}}\\\\\displaystyle{\sec \theta = \sf \dfrac{hypotenuse}{adjacent} = \dfrac{r}{x}}\\\\\displaystyle{\cot \theta = \sf \dfrac{adjacent}{opposite} = \dfrac{x}{y}}[/tex]

Since we are given tangent relation, we know that [tex]\displaystyle{y = 2ab}[/tex] and [tex]\displaystyle{x = a^2-b^2}[/tex], all we have to do is to find hypotenuse or radius (r) which you can find by applying Pythagoras Theorem.

[tex]\displaystyle{r=\sqrt{x^2+y^2}}[/tex]

Therefore:

[tex]\displaystyle{r=\sqrt{(a^2-b^2)^2+(2ab)^2}}\\\\\displaystyle{r=\sqrt{a^4-2a^2b^2+b^4+4a^2b^2}}\\\\\displaystyle{r=\sqrt{a^4+2a^2b^2+b^4}}\\\\\displaystyle{r=\sqrt{(a^2+b^2)^2}}\\\\\displaystyle{r=a^2+b^2}[/tex]

Now we can find other trigonometric ratios by simply substituting the given information below:

Hence:

[tex]\displaystyle{\sin \theta = \dfrac{y}{r} = \dfrac{2ab}{a^2+b^2}}\\\\\displaystyle{\cos \theta = \dfrac{x}{r} = \dfrac{a^2-b^2}{a^2+b^2}}\\\\\displaystyle{\csc \theta = \dfrac{r}{y} = \dfrac{a^2+b^2}{2ab}}\\\\\displaystyle{\sec \theta = \dfrac{r}{x} = \dfrac{a^2+b^2}{a^2-b^2}}\\\\\displaystyle{\cot \theta = \dfrac{x}{y} = \dfrac{a^2-b^2}{2ab}}[/tex]

will be other trigonometric ratios.

Best guess for the function is

[tex]\displaystyle f(x) = \sum_{n=1}^\infty \frac{x^n}{n^2}[/tex]

By the ratio test, the series converges for

[tex]\displaystyle \lim_{n\to\infty} \left|\frac{x^{n+1}}{(n+1)^2} \cdot \frac{n^2}{x^n}\right| = |x| \lim_{n\to\infty} \frac{n^2}{(n+1)^2} = |x| < 1[/tex]

When [tex]x=1[/tex], [tex]f(x)[/tex] is a convergent [tex]p[/tex]-series.

When [tex]x=-1[/tex], [tex]f(x)[/tex] is a convergent alternating series.

So, the interval of convergence for [tex]f(x)[/tex] is the closed interval [tex]\boxed{-1 \le x \le 1}[/tex].

The derivative of [tex]f[/tex] is the series

[tex]\displaystyle f'(x) = \sum_{n=1}^\infty \frac{nx^{n-1}}{n^2} = \frac1x \sum_{n=1}^\infty \frac{x^n}n[/tex]

which also converges for [tex]|x|<1[/tex] by the ratio test:

[tex]\displaystyle \lim_{n\to\infty} \left|\frac{x^{n+1}}{n+1} \cdot \frac n{x^n}\right| = |x| \lim_{n\to\infty} \frac{n}{n+1} = |x| < 1[/tex]

When [tex]x=1[/tex], [tex]f'(x)[/tex] becomes the divergent harmonic series.

When [tex]x=-1[/tex], [tex]f'(x)[/tex] is a convergent alternating series.

The interval of convergence for [tex]f'(x)[/tex] is then the closed-open interval [tex]\boxed{-1 \le x < 1}[/tex].

Differentiating [tex]f[/tex] once more gives the series

[tex]\displaystyle f''(x) = \sum_{n=1}^\infty \frac{n(n-1)x^{n-2}}{n^2} = \frac1{x^2} \sum_{n=1}^\infty \frac{(n-1)x^n}{n} = \frac1{x^2} \left(\sum_{n=1}^\infty x^n - \sum_{n=1}^\infty \frac{x^n}n\right)[/tex]

The first series is geometric and converges for [tex]|x|<1[/tex], endpoints not included.

The second series is [tex]f'(x)[/tex], which we know converges for [tex]-1\le x<1[/tex].

Putting these intervals together, we see that [tex]f''(x)[/tex] converges only on the open interval [tex]\boxed{-1 < x < 1}[/tex].

The number which belongs to the set of rational numbers and the set of integers is -115 which is third option,-15 which is fourth option.

Given four options:

We are required to find the number which is included in the set of rational numbers and the set of integers.

Rational numbers are those numbers which can be written in the form of p/q in which q cannot be equal to zero because if q becomes zero then the fraction becomes infinity.

-5.5 is not a rational number,

-0.5 is also not a rational number.

-115 is a rational number and also an integer.

-15 is a rationalnumber and also an integer.

Hence the number which belongs to the set of rational numbers and the set of integers is -115 which is third option,-15 which is fourth option.

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We need at least 7 terms of the Maclaurin series for ln(1 + x)  to estimate ln 1.4 to within 0.0001

For given question,

We have been given a function f(x) = ln(1 + x)

We need to find  the estimate of In(1.4) within 0.001 by applying the function of the Maclaurin series for f(x) = In (1 + x)

The expansion of ln(1 + x) about zero is:

[tex]ln(1+x)=x-\frac{x^2}{2} + \frac{x^3}{3} -\frac{x^4}{4} +\frac{x^5}{5} -\frac{x^6}{6} +.~.~.[/tex]

where -1 ≤ x ≤ 1

To estimate the value of In(1.4), let's replace x with 0.4

[tex]\Rightarrow ln(1+0.4)=0.4-\frac{0.4^2}{2} + \frac{0.4^3}{3} -\frac{0.4^4}{4} +\frac{0.4^5}{5} -\frac{0.4^6}{6} +.~.~.[/tex]

From the above calculations, we will realize that the value of  [tex]\frac{0.4^5}{5}=0.002048[/tex] and [tex]\frac{0.4^6}{6}=0.000683[/tex]  which are approximately equal to 0.001

Hence, the estimate of In(1.4) to the term [tex]\frac{0.4^6}{6}[/tex]  is enough to justify our claim.

Therefore,  we need at least 7 terms of the Maclaurin series for function ln(1 + x)  to estimate ln 1.4 to within 0.0001

Learn more about the Maclaurin series here:

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Answer:

Separating the 3x and -5 apart from the (x+1)

Step-by-step explanation:

It would turn out to be (3x-5)(x+1) !

Factor out x+1 from the expression

(x+1) x (3x-5)

Answer:

9/4

Step-by-step explanation:

We follow bodmas

4/3 +( -1/6 + 13/12)

( lcm = 12)

( -2+ 13/12)

( 11/ 12)

4/3 + ( 11/12)

4/3 + 11/12

lcm = 12 also

and that will equal to

=16 + 11/ 12

= 27/ 12

divide by 3 to simplest form

= 9/4

Answer:

x = 5, y =2

Step-by-step explanation:

I guess the question is saying x+y = 7 and 3x-2y = 11?

then there are multiple ways but

I will multiply the first one by 2 so 2x+2y = 14

you add the equations to get 5x = 25 so x = 5 plug x into the first equation you get y = 2

if that isn't what the question means just comment and I'll change it

Answer:

2

Step-by-step explanation:

given x=1 and y=7

now, given expression ,

x+y/4

by putting the values of the x and y ,we get

x+y/4

= 1+7/4

= 8/4

= 2 (Ans.)