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luda_lava [24]
2 years ago
5

Solve for y please help

Mathematics
1 answer:
ss7ja [257]2 years ago
7 0

Answer:

The answer is (-16.6)

Step-by-step explanation:

y/2 ≥ -8.3

y ≥ (-8.3) × 2

y ≥ -16.6

Thus, The value of y is (-16.6)

<u>-TheUnknownScientist 72</u>

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Y=3(50)x
solmaris [256]

Answer:

y=3(50)×

y=3(50) 8

50×3=150

150×8=1,200

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2 years ago
The dot plots show the heights of the players on two basketball teams.
astra-53 [7]

Answer:

there is the dot plot plz show me

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
1. Which is NOT a solution to the inequality below?<br> 3x + 13 ≥ -8<br> -8<br> -7<br> -6<br> -5
Viktor [21]

Answer:

-8 is not a solution to the inequality

Step-by-step explanation:

3x + 13 \geqslant  - 8 \\ 3x \geqslant  - 21 \\ x \geqslant  - 7 \\

6 0
3 years ago
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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
Jada's dog ran 20 laps around the track and then a quarter of that again. How far did Jada's dog run in all? *
Alenkasestr [34]

Answer:

25 laps

Step-by-step explanation:

The dog ran 20 laps and a quarter of 20. 20 divided by 4 equals 5 so it would be 20+5 which equals 25

6 0
3 years ago
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