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2 years ago

Pls answer Math-iready

Mathematics

1 answer:

Ymorist [56]2 years ago

4 0

Answer:

1 - Variable

2 - Corfficient

3 - Term

4 - Constant

Step-by-step explanation:

So, I am pretty srue the way the boxes are now is the correct way already. (I should be wrong?)

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Which set is closed under the operation multiplication?

vlada-n [284]

D is under multiplication

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2 years ago

Find all solutions of the given system of equations and check your answer graphically. HINT [See Examples 1-4.] (If there is no

Likurg_2 [28]

Answer:

$Infinitely\ many\ solutions\ exist.\\\\Solutions\ are\ (x,\frac{3}{4}x-1)$

Step-by-step explanation:

$Given\ equations\ are\\\\3x-4y=4.................eq(1)\\\\9x-12y=12..............eq(2)\\\\divide\ eq(2)\ by\ 3\\\\\frac{1}{3}(9x-12y=12)\\\\\Rightarrow 3x-4y=4\\\\Hence\ equations\ represent\ the\ same\ line.\\Hence\ Infinitely\ many\ solutions\ exist.\\\\3x-4y=4\\\\4y=3x-4\\\\y=\frac{3}{4}x-1\\\\Solutions\ are\ (x,\frac{3}{4}x-1)$

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2 years ago

-3 (5 g +1) + 8 g

Komok [63]

Answer:

-7g-3

Step-by-step explanation:

in order to evaluate for this mathematical expression, we need to consider the negative sign carefully and then open the bracket then evaluate.

<u>solutuion</u>

-3 (5 g +1) + 8 g

-15g -3 + 8g

-15g +8g -3

-7g-3

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2 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.