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Natali [406]
3 years ago
7

Simplify 7891 * 1/2 please help

Mathematics
2 answers:
Julli [10]3 years ago
8 0
The answer to this question is:

3945 1/2

Hope this helps you Alan :)
enot [183]3 years ago
6 0
Expression: = 7891 * 1/2

= 3945.5

In short, Your Answer would be: 3945.5

Hope this helps!
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Write the number 16,107,320 in expanded form
yarga [219]
16,107,320=10,000,000+6,000,000+100,000+7,000+300+20


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3 years ago
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Please help i don't understand this​
tatyana61 [14]

Answer:

A. x = (k + c)/4

Step-by-step explanation:

In this question, they're asking for what x equals in terms of k.

You know that 4x - c = k

add c on both sides, you get:

4x = k + c

divide 4 on both sides to isolate the x variable, you get

x = (k + c)/4

<em>It's a simple question when you think about it! If this helped, please mark it brainliest.....thank you....................</em>

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3 years ago
Given: PRbisects ZQPS, PQ = 12units, and PS= 12units
Ganezh [65]

Answer:

Use SAS to show that triangles PRQ and PRS are congruent.

Step-by-step explanation:

Since PR bisects angle QPS, angles QPR and SPR are congruent. By reflexive property of congruence, PR is congruent to itself. Since PQ is congruent to PS, we can use SAS to show that the two triangles are congruent. By CPCTC, QR is congruent to SR.

3 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.

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Ms.Diaz wants to divide her class of 30 students into equal-sized groups.What are her choices?
RoseWind [281]
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