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3 years ago

A house purchased for $226,000 has lost 4% of its value each year for the past five years. What is it worth now?

Mathematics

1 answer:

SSSSS [86.1K]3 years ago

7 0

Answer:

The answer would be $184,274.23.

Step-by-step explanation:

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If line A contains Q(5,1) and is parallel to line MN with M(-2,4) and N(2,1),which ordered pair would be on the perpendicular tr

pashok25 [27]

Answer:

can u explain more?

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

If ΔABC ≅ ΔDEF by SSS and AB = 5x + 6 and DE = 3x +10, then what must be the value of x?

ANTONII [103]

Answer:

ab=de corresponding side of the congruent triangle are equal

5x+6=3x+10

5x-3x=10-6

2x=4

x=4/2=2

2 is your answer

4 0

3 years ago

Given the net of the rectangular prism, what is its surface area?

Llana [10]

Answer:

Step-by-step explanation:

Look at the picture.

We have :

4 rectangles 3m × 4m → A₁ = 3 · 4 = 12 m²

2 rectangles 3m × 3m → A₂ = 3 · 3 = 9 m²

Therefore the Surface Area of the rectangular prism is:

S.A. = 4A₁ + 2A₂

Substitute:

S.A. = 4(12) + 2(9) = 48 + 18 = 66

5 0

3 years ago

Find the taylor series centered at the given value of a and find the associated radius of convergence. (1) f(x) = 1 x , a = 1 (2

Kitty [74]

a) The radius of convergence is calculated as

R=1.

b) Due to the fact that it converges in every direction, the radius of convergence is either infinity or zero.

<h3>What is the associated radius of convergence.?</h3>

(a)

Take into consideration the function f with respect to the number a,

$f(x)=\frac{1}{x}, \quad a=1$

In case you forgot, the Taylor series for the function $f$ at the number a looks like this:

$\begin{aligned}f(x) &=\sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n !}(x-a)^{n} \\&=f(a)+\frac{f^{\prime}(a)}{1 !}(x-a)+\frac{f^{*}(a)}{2 !}(x-a)^{2}+\ldots\end{aligned}$

Determine the function f as well as any derivatives of the function $f by setting a=1 and working backward from there.

$\begin{aligned}f(x) &=\frac{1}{x} & f(1)=\frac{1}{1}=1 \\\\f^{\prime}(x) &=-\frac{1}{x^{2}} & f^{\prime}(1)=-\frac{1}{(1)^{2}}=-1 \\\\f^{\prime \prime}(x) &=\frac{2}{x^{3}} & f^{\prime \prime}(1)=\frac{2}{(1)^{3}}=2 \\\\f^{\prime \prime}(x) &=-\frac{2 \cdot 3}{x^{4}} & f^{\prime \prime}(1)=-\frac{2 \cdot 3}{(1)^{4}}=-2 \cdot 3 \\\\f^{(*)}(x) &=\frac{2 \cdot 3 \cdot 4}{x^{5}} & f^{(n)}(1)=\frac{2 \cdot 3 \cdot 4}{(1)^{5}}=2 \cdot 3 \cdot 4\end{aligned}$

At the point when a = 1, the Taylor series for the function f looks like this:

$f(x) &=f(a)+\frac{f^{\prime}(a)}{1 !}(x-a)+\frac{f^{\prime \prime}(a)}{2 !}(x-a)^{2}+\frac{f^{\prime \prime \prime}(a)}{3 !}(x-a)^{3}+\cdots \\\\&=f(1)+\frac{f^{\prime}(1)}{1 !}(x-1)+\frac{f^{\prime}(1)}{2 !}(x-1)^{2}+\frac{f^{\prime \prime}(1)}{3 !}(x-1)^{3}+\cdots \\$

$&=1+\frac{-1}{1 !}(x-1)+\frac{2}{2 !}(x-1)^{2}+\frac{-2 \cdot 3}{3 !}(x-1)^{3}+\frac{2 \cdot 3 \cdot 4}{4 !}(x-1)^{4}+\cdots \\\\&=1-(x-1)+(x-1)^{2}-(x-1)^{3}+(x-1)^{4}+\cdots \\\\&=\sum_{1=0}^{\infty}(-1)^{n}(x-1)^{n}$

In conclusion,

$&=\sum_{1=0}^{\infty}(-1)^{n}(x-1)^{n}$

Find the radius of convergence by using the Ratio Test in the following manner:

$\begin{aligned}L &=\lim _{n \rightarrow \infty}\left|\frac{a_{n+1}}{a_{n}}\right| \\&=\lim _{n \rightarrow \infty} \frac{(-1)^{n+1}(x-1)^{n+1}}{(-1)^{n}(x-1)^{n}} \mid \\&=\lim _{n \rightarrow \infty}|x-1| \\&=|x-1|\end{aligned}$

The convergence of the series when L<1, that is, |x-1|<1.

The radius of convergence is calculated as

R=1.

For B

Take into consideration the function f with respect to the number a,

$a_{n}=(-1)^{n}(x-1)^{n}$

$f(x)=\left(x^{2}+2 x\right) e^{x}, a=0$ The Taylor series for f(x)=e^{x} at a=0 is,

$e^{2}=1+x+\frac{x^{2}}{2 !}+\frac{x^{3}}{3 !}+\frac{x^{4}}{4 !}+\ldots$

$f(x) &=\left(x^{2}+2 x\right) e^{x} \\&=\left(x^{2}+2 x\right)\left(1+x+\frac{x^{2}}{2 !}+\frac{x^{3}}{3 !}+\frac{x^{4}}{4 !}+\ldots\right)+2 x\left(1+x+\frac{x^{2}}{2 !}+\frac{x^{3}}{3 !}+\frac{x^{4}}{4 !}+\ldots\right) \\&=x^{2}\left(1+x+\frac{x^{2}}{2 !}+\frac{x^{3}}{3 !}+\frac{x^{4}}{4 !}+\ldots\right)+\left(\frac{x^{4}}{2 !}+\frac{x^{5}}{3 !}+\frac{x^{6}}{4 !}+\ldots\right)+\left(2 x+2 x^{2}+\frac{2 x^{3}}{2 !}+\frac{2 x^{4}}{3 !}+\frac{2 x^{5}}{4 !}+\ldots\right) \\$

$&=\left(x^{2}+x^{3}+\frac{x^{4}}{2 !}\right) \\&=2 x+3 x^{2}+\left(1+\frac{2}{2 !}\right) x^{3}+\left(\frac{1}{2 !}+\frac{2}{3 !}\right) x^{4}+\left(\frac{1}{4 !}\right) x^{5}+\ldots \\&=2 x+3 x^{2}+2 x^{3}+\frac{5}{6} x^{4}+\frac{1}{4} x^{5}+\ldots$

Due to the fact that it converges in every direction, the radius of convergence is either infinity or zero.

Read more about convergence

brainly.com/question/15415793

#SPJ4

The complete question is attached below

3 0

2 years ago

Five students, Alec, Bryn, Igor, Molly and Suzy, too! en exam The exern was marked out of 30.

emmainna [20.7K]

Answer:

There are 3 possibilities:

$\begin{tabular}{lccc}Alec&7&7&8\\Bryn&18&9&7\\Igor&21&21&24\\Molly&26&17&15\\Suzy&8&26&26\end{tabular}$

Step-by-step explanation:

This logic puzzle makes use of the relations between numbers and their mean.

If two scores in the range of 7 to 26 have a ratio of exactly 3, they must be either 7 and 21, or 8 and 24. In order for the mean to be exactly 16, the total of the 5 scores must be 5×16 = 80.

<h3>one solution</h3>

If the 3:1 range involves scores of 21 and 7, then we know ...

Alec got 7

Igor got 21

The total of these and the score of 26 is 54. The remaining two scores must total 80-54 = 26. We know that there must be two scores that have a difference of 8, so there are two possibilities:

Molly got 26 and Bryn got 18, leaving Suzy with a score of 8.

Molly got 17 and Bryn got 9, leaving Suzy with a score of 26.

The possibilities here are ...

Alec 7, Bryn 18, Igor 21, Molly 26, Suzy 8

Alec 7, Bryn 9, Igor 21, Molly 17, Suzy 26

<h3>another solution</h3>

If the 3:1 range involves scores of 24 and 8, then we know ...

Alec got 8

Igor got 24

The total of these and the score of 26 is 58. The remaining two scores must total 80-58 = 22. That two scores have a difference of 8 leaves open one possibility: the remaining two scores are 7 and 15. (If one were 18=26-8, the other would be 22-18=4, too low for the given range.)

Alec 8, Bryn 7, Igor 24, Molly 15, Suzy 26

___

Possible marks are ...

$\begin{tabular}{lccc}Alec&7&7&8\\Bryn&18&9&7\\Igor&21&21&24\\Molly&26&17&15\\Suzy&8&26&26\end{tabular}$

_____
<em>Additional comment</em>

When two numbers are required to have a specific sum (s) and a specific difference (d), the two numbers are (s+d)/2 and (s-d)/2.

5 0

2 years ago