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Murrr4er [49]
3 years ago
13

What two decimals are equivalent to 3.300​

Mathematics
1 answer:
Rashid [163]3 years ago
4 0
3.3 and 3.30 are equivalent to 3.300
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Charlie was carrying around the steps to solve Square root of 6x plus 4 = 8. On his way through the Math Factory, Charlie droppe
fomenos

Answer: Order will be F,D,C and Fourth Option is correct and  x = 10

Step-by-step explanation:

Since we have given that

\sqrt{6x+4}=8

We first transpose the square root to the right , so it becomes square of 8,i.e.

6x+4=8^2\\\\6x+4=64

Now, transpose 4 to the right so it will get subtract from 64 i.e.

6x=64-4\\\\6x=60

Since 6 is multiplied to x on tranposing it will get divided by 60 i.e.

x=\frac{60}{6}\\\\x=10

Hence, on simplification, we get x=10.

Hence , the order is F,D,C.

6 0
3 years ago
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Which statement must be true about line TU?
just olya [345]
Line tu is parallel to rs
5 0
3 years ago
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The side lengths of a 45-45-90 triangle are in the ratio 1 : 1: V2. What is tan 45°?
Artyom0805 [142]

Answer:

B. 1/2

Step-by-step explanation:

divide/

3 0
3 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
What's a proportion of 15/3
erastova [34]

Answer:

5/1 or 5

Step-by-step explanation:

i hope this helps :)

3 0
2 years ago
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