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3 years ago

1862 divided by 17383

Mathematics

1 answer:

Kipish [7]3 years ago

6 0

Answer:

do i round to the nearest hundredth?if so it would be 0.11

Step-by-step explanation:

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The mean age of 5 people in a room is 27 years.

bija089 [108]

The answer is 135
27 x 5=135

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3 years ago

Typical misalignments between product design and operations include A. technology, reward systems, and infrastructure. B. techno

Lady bird [3.3K]

Answer:

(A) technology, reward systems, and infrastructure.

Explanation:

Misalignments can occur in technology, infrastructure, and reward systems, and also in marketing. A company is misaligned when people pursue goals and agendas that are incongruent with each other and do not combine to effectively advance a single purpose.

In fact, in marketing, misalignment is the scourge of the modern marketing department, and it’s no surprise why there is so many channels to manage with no way to predict the customer journey. Highly fragmented roles and teams executing on compressed time cycles. Intense pressure to drive business impact, while staff are drowning in disconnected data and unclear strategy.

4 0

3 years ago

9. Two points in the plane, A(−3,8) and ????(17,8), represent the endpoints of the diameter of a circle.

kherson [118]

Answer:

a) (7,8)

b) r = 10

c) $(x-7)^2+(y-8)^2=100$

Step-by-step explanation:

a) Given the endpoints the diameter of the circle as A(-3,8) and B(17,8)

we should realize that the center of the circle lies exactly at the midpoint of these two points. let's denote the midpoint as $(x_m,y_m)$

$(x_m, y_m) = \left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\right)$

$(x_m, y_m) = \left(\dfrac{-3+17}{2},\dfrac{8+8}{2}\right)$

$(x_m, y_m) = (7,8)$

this is the coordinate of the center of the circle.

b) The radius of the circle can be easily found by using the distance formula between the center point and either of the two endpoints of the diameter.

the general distance formula is:

$r = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

in our case the equation changes to: (selecting the centre and point B)

$r = \sqrt{(x_m-x_2)^2+(y_m-y_2)^2}$

$r = \sqrt{(7-17)^2+(8-8)^2}$

$r = 10$

side note: there are more alternate ways to find the radius for e.g (you can use the distance formula between the points A and B and that'll give you the length of the diameter which you can divide by 2 to get the radius) OR (you don't need to use the distance formula at all <u>since in this particular case</u> all the coordinates lie on the same horizontal line, so by simply subtracting the two x-coordinates of the center and either of A or B)

c) the ingredients needed to make the equation of the circle are the

coordinates of the center: $(x_m,y_m) = (7,8)$
radius of the circle: $r = 10$

we can put this in the formula of the circle:

$(x-a)^2+(y-b)^2=r^2$

in our case the equation changes to:

$(x-x_m)^2+(y-y_m)^2=r^2$

$(x-7)^2+(y-8)^2=(10)^2$

$(x-7)^2+(y-8)^2=100$

this is the equation of the circle!

6 0

3 years ago

What are 2 binomials that are factors of this trinomial? x^2-x-20

liberstina [14]

Answer:

(x-5) (x+4)

Step-by-step explanation:

x^2 -x -20

What two terms multiply to -20 and add to -1

-5*4 = -20

-5+4 = -1

(x-5) (x+4)

8 0

3 years ago

Read 2 more answers

Taylor Series Questions!

riadik2000 [5.3K]

5.
$f(x)=\sin x\implies f(\pi)=0$
$f'(x)=\cos x\implies f'(\pi)=-1$
$f''(x)=-\sin x\implies f''(\pi)=0$
$f'''(x)=-\cos x\implies f'''(\pi)=1$

Clearly, each even-order derivative will vanish, and the terms that remain will alternate in sign, so the Taylor series is given by

$f(x)=-(x-\pi)+\dfrac{(x-\pi)^3}{3!}-\dfrac{(x-\pi)^5}{5!}+\cdots$
$f(x)=\displaystyle\sum_{n\ge0}\frac{(-1)^{n-1}(x-\pi)^{2n+1}}{(2n+1)!}$

Your answer is off by a sign - the source of this error is the fact that you used the series expansion centered at $x=0$ , not $x=\pi$ , and so the sign on each derivative at $x=\pi$ is opposite of what it should be. I'm sure you can figure out the radius of convergence from here.

- - -

6. Note that this is already a polynomial, so the Taylor series will strongly resemble this and will consist of a finite number of terms. You can get the series by evaluating the derivatives at the given point, or you can simply rewrite the polynomial in $x$ as a polynomial in $x-2$ .

$f(x)=x^6-x^4+2\implies f(2)=50$
$f'(x)=6x^5-4x^3\implies f'(2)=160$
$f''(x)=30x^4-12x^2\implies f''(2)=432$
$f'''(x)=120x^3-24x\implies f'''(2)=912$
$f^{(4)}(x)=360x^2-24\implies f^{(4)}(2)=1416$
$f^{(5)}(x)=720x\implies f^{(5)}(2)=1440$
$f^{(6)}(x)=720\implies f^{(6)}(2)=720$
$f^{(n\ge7)}(x)=0\implies f^{(n\ge7)}(2)=0$

$\implies f(x)=50+160(x-2)+216(x-2)^2+152(x-2)^3+59(x-2)^4+12(x-2)^5+(x-2)^6$

If you expand this, you will end up with $f(x)$ again, so the Taylor series must converge everywhere.

I'll outline the second method. The idea is to find coefficients so that the right hand side below matches the original polynomial:

$x^6-x^4+2=(x-2)^6+a_5(x-2)^5+a_4(x-2)^4+a_3(x-2)^3+a_2(x-2)^2+a_1(x-2)+a_0$

You would expand the right side, match up the coefficients for the same-power terms on the left, then solve the linear system that comes out of that. You would end up with the same result as with the standard derivative method, though perhaps more work than necessary.

- - -

7. It would help to write the square root as a rational power first:

$f(x)=\sqrt x=x^{1/2}\implies f(4)=2$
$f'(x)=\dfrac{(-1)^0}{2^1}x^{-1/2}\implies f'(4)=\dfrac1{2^2}$
$f''(x)=\dfrac{(-1)^1}{2^2}x^{-3/2}\implies f''(4)=-\dfrac1{2^5}$
$f'''(x)=\dfrac{(-1)^2(1\times3)}{2^3}x^{-5/2}\implies f'''(4)=\dfrac3{2^8}$
$f^{(4)}(x)=\dfrac{(-1)^3(1\times3\times5)}{2^4}x^{-7/2}\implies f^{(4)}(4)=-\dfrac{15}{2^{11}}$
$f^{(5)}(x)=\dfrac{(-1)^4(1\times3\times5\times7)}{2^5}x^{-9/2}\implies f^{(5)}(4)=\dfrac{105}{2^{14}}$

The pattern should be fairly easy to see.

$f(x)=2+\dfrac{x-4}{2^2}-\dfrac{(x-4)^2}{2^5\times2!}+\dfrac{3(x-4)^3}{2^8\times3!}-\dfrac{15(x-4)^4}{2^{11}\times4!}+\cdots$
$f(x)=2+\displaystyle\sum_{n\ge1}\dfrac{(-1)^n(-1\times1\times3\times5\times\cdots\times(2n-3)}{2^{3n-1}n!}(x-4)^n$

By the ratio test, the series converges if

$\displaystyle\lim_{n\to\infty}\left|\frac{\dfrac{(-1)^{n+1}(-1\times\cdots\times(2n-3)\times(2n-1))(x-4)^{n+1}}{2^{3n+2}(n+1)!}}{\dfrac{(-1)^n(-1\times\cdots\tiems(2n-3))(x-4)^n}{2^{3n-1}n!}}\right|$
$\implies\displaystyle\frac{|x-4|}8\lim_{n\to\infty}\frac{2n-1}{n+1}=\frac{|x-4|}4$
$\implies |x-4|$

so that the ROC is 4.

- - -

10. Without going into much detail, you should have as your Taylor polynomial

$\sin x\approx T_4(x)=\dfrac12+\dfrac{\sqrt3}2\left(x-\dfrac\pi6\right)-\dfrac14\left(x-\dfrac\pi6\right)^2-\dfrac1{4\sqrt3}\left(x-\dfrac\pi6\right)^3+\dfrac1{48}\left(x-\dfrac\pi6\right)^4$

Taylor's inequality then asserts that the error of approximation on the interval $0\le x\le\dfrac\pi3$ is given by

$|\sin x-T_4(x)|=|R_4(x)|\le\dfrac{M\left|x-\frac\pi6\right|^5}{5!}$

where $M$ satisfies $|f^{(5)}(x)|\le M$ on the interval.

We know that $(\sin x)^{(5)}=\cos x$ is bounded between -1 and 1, so we know $M=1$ will suffice. Over the given interval, we have $\left|x-\dfrac\pi6\right|\le\dfrac\pi6$ , so the remainder will be bounded above by

$|R_4(x)|\le\dfrac{1\times\left(\frac\pi6\right)^5}{5!}=\dfrac{\pi^5}{933120}\approx0.000328$

which is to say, over the interval $0\le x\le\dfrac\pi3$ , the fourth degree Taylor polynomial approximates the value of $\sin x$ near $x=\dfrac\pi6$ to within 0.000328.

7 0

4 years ago