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Alchen [17]
3 years ago
13

X=y/20-1/4 solve for y and show properties

Mathematics
1 answer:
andriy [413]3 years ago
3 0

Answer:

y = 20x + 5

Step-by-step explanation:

X =  Y/20 - 1/4

In this type of expression you will first of all find the LCM of the right hand side

x = y - 5/ 20

the next thing to do is to cross multiply

we have,

20x = y - 5

dont forget we are to solve for y in the equation, so to solve for y we have

20x + 5 = y

therefore y will be expressed in terms of x as written below

y = 20x + 5

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Juno calculated the area of a square to be 4/9 square yard. Which shows the side length of the square?
Stella [2.4K]
\bf \textit{area of a square}\\\\
A=s^2~~
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s=side's~length\\
--------\\
A=\frac{4}{9}
\end{cases}\implies \cfrac{4}{9}=s^2
\\\\\\
\sqrt{\cfrac{4}{9}}=s\implies \cfrac{\sqrt{4}}{\sqrt{9}}=s\implies \cfrac{2}{3}=s
3 0
3 years ago
I will mark BRAINLIEST if your answer is accurate!
Mila [183]

Answer:

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Step-by-step explanation:

7 0
3 years ago
Which of the following expressions is a factor of the polynomial x 2 +3/2x-1
Elodia [21]

Answer:

\large \boxed{\sf \ \ \ (x+2)(x-\dfrac{1}{2}) \ \ \ }

Step-by-step explanation:

Hello,

let's solve

x^2+\dfrac{3}{2}x-1=0\\\\ 2x^2+3x-2=0 \ \text{multiply by 2}\\\\\\

\Delta=b^2-4ac=9+4*2*2=9+16=25

There are two solutions

x_1=\dfrac{-3-\sqrt{25}}{4}\\\\x_1=\dfrac{-3-5}{4}\\\\x_1=\dfrac{-8}{4}\\\\\boxed{x_1=-2}

And

x_2=\dfrac{-3+\sqrt{25}}{4}\\\\x_2=\dfrac{-3+5}{4}\\\\x_2=\dfrac{2}{4}\\\\\boxed{x_2=\dfrac{1}{2}}

So we can write

x^2+\dfrac{3}{2}x-1\\\\=(x-x_1)(x-x_2)\\\\=(x+2)(x-\dfrac{1}{2})

Hope this helps.

Do not hesitate if you need further explanation.

Thank you

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3 years ago
Write rational number in number line -7 upon 2​
Leno4ka [110]

Answer:

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Step-by-step explanation:

7 0
2 years ago
Find the Taylor series for f(x) centered at the given value of a. [Assume that f has a power series expansion. Do not show that
FromTheMoon [43]

Answer:

The Taylor series is \ln(x) = \ln 3 + \sum_{n=1}^{\infty} (-1)^{n+1} \frac{(x-3)^n}{3^n n}.

The radius of convergence is R=3.

Step-by-step explanation:

<em>The Taylor expansion.</em>

Recall that as we want the Taylor series centered at a=3 its expression is given in powers of (x-3). With this in mind we need to do some transformations with the goal to obtain the asked Taylor series from the Taylor expansion of \ln(1+x).

Then,

\ln(x) = \ln(x-3+3) = \ln(3(\frac{x-3}{3} + 1 )) = \ln 3 + \ln(1 + \frac{x-3}{3}).

Now, in order to make a more compact notation write \frac{x-3}{3}=y. Thus, the above expression becomes

\ln(x) = \ln 3 + \ln(1+y).

Notice that, if x is very close from 3, then y is very close from 0. Then, we can use the Taylor expansion of the logarithm. Hence,  

\ln(x) = \ln 3 + \ln(1+y) = \ln 3 + \sum_{n=1}^{\infty} (-1)^{n+1} \frac{y^n}{n}.

Now, substitute \frac{x-3}{3}=y in the previous equality. Thus,

\ln(x) = \ln 3 + \sum_{n=1}^{\infty} (-1)^{n+1} \frac{(x-3)^n}{3^n n}.

<em>Radius of convergence.</em>

We find the radius of convergence with the Cauchy-Hadamard formula:

R^{-1} = \lim_{n\rightarrow\infty} \sqrt[n]{|a_n|},

Where a_n stands for the coefficients of the Taylor series and R for the radius of convergence.

In this case the coefficients of the Taylor series are

a_n = \frac{(-1)^{n+1}}{ n3^n}

and in consequence |a_n| = \frac{1}{3^nn}. Then,

\sqrt[n]{|a_n|} = \sqrt[n]{\frac{1}{3^nn}}

Applying the properties of roots

\sqrt[n]{|a_n|} = \frac{1}{3\sqrt[n]{n}}.

Hence,

R^{-1} = \lim_{n\rightarrow\infty} \frac{1}{3\sqrt[n]{n}} =\frac{1}{3}

Recall that

\lim_{n\rightarrow\infty} \sqrt[n]{n}=1.

So, as R^{-1}=\frac{1}{3} we get that R=3.

8 0
3 years ago
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