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2 years ago

A person has a bag of coins.

Mathematics

2 answers:

lbvjy [14]2 years ago

8 0

Answer:

$15.6

Step-by-step explanation:

First, let's assume that x is the total money in the bag of coins.

1. x = total money

4/13 of the coins are pennies, so 4/13 of x equals to the number of pennies.

4/13(x) = P (pennies)

What is left that is not pennies is 9/13(x). And out of these, 1/4 are nickels. Which means:

1/4(9/13[x]) = N (nickels)

If 1/4 of the remaining are nickels, then 3/4 are dimes. So:

3/4(9/13[x]) = $8.10 (dimes' value)

After calculating the one with the value:

27/52(x) = $8.10

27x = $8.10 times 52

x = ($8.10 times 52)/27

x = 52 x $0.3

x = $15.6

Ans. The value of the bag of coins is $15.6

Y_Kistochka [10]2 years ago

5 0

The answer is $15.06

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A door to a playhouse is 50 inches tall. Which of the following is another measure equal to the height?

Effectus [21]

Answer:

4 ft 2 in.

Step-by-step explanation:

A door to a playhouse is 50 inches tall

we know that

1 in---------------> is 0.0833333 ft

50 in-------------> 50*0.0833333---------> 4.167 ft

4.167 ft----------> [4 ft+0.167 ft]

remember that

1 in---------------> is 0.0833333 ft

X in----------------> 0.167 ft

X=0.167/0.0833333----------> X=2 in

4.167 ft----------> [4 ft+0.167 ft]--------> [4 ft+2 in]

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2 years ago

Estimate the area of the rectangle. A)8 B)10 C)14 D)20

seraphim [82]

Answer:

A or B, I think it's A...

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3 years ago

Bab can’t do quick mafs. Help?

iragen [17]

Answer:

(5, -3)

Step-by-step explanation:

Midpoint formula:

$(x,y ) = (\frac{x_{1}+x_{2} }{2} ,\frac{y_{1} +y_{2} }{2})\\= (\frac{3+7}{2}, \frac{-1+(-5)}{2} )\\=(\frac{10}{2}, \frac{-6}{2} )\\= (5,-3)$

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3 years ago

Find the base length of a shipping box in the shape of a triangular prism the shipping box has a volume of 276 cubic feet a base

Luba_88 [7]

The volume is given by:
V = A * H
Where,
A: area of the base
H: prism height
Clearing the area we have:
A = V / H
A = 276/10
A = 27.6 feet ^ 2
The area of the base is given by:
A = (1/2) * (b) * (h)
Where,
b: base of the triangle
h: height of the triangle
Clearing b we have:
b = (2 * A) / (h)
Substituting values:
b = (2 * 27.6) / (6.9)
b = 8 feet
Answer:
the base length is:
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3 years ago

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.

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3 years ago