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3 years ago

Is 14.875 rational or irrational

Mathematics

2 answers:

zaharov [31]3 years ago

4 0

Answer:

Rational

Step-by-step explanation:

119/8 is it in fractional form.

Delicious77 [7]3 years ago

3 0

Answer:

not rational

Step-by-step explanation:

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xenn [34]

How do you find the area of a parallelogram?
Area=base * height
S0, you multiple 10in by 6in to get 60 inches squared. Hope this helped!

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3 years ago

Read 2 more answers

-2 - (8) I don’t know what is the answer can someone help me

MrMuchimi

Answer:

its 16

Step-by-step explanation:

because if you take off the parenthathesis you get -2-8 then you multiply it and since they are both negative you get a positive, so that means you get positive 16

4 0

4 years ago

Which shows the expression below simplified? 0.00048 - (3.4 × 10-5) A. 4.46 × 10-4 B. 3.39952 × 10-5 C. 4.46 × 10-5 D. 1.4 × 10-

MakcuM [25]

Option C

The simplified form of given expression is $4.46 \times 10^{-4}$

Solution:

Given that we have to simplify the expression shown below

$0.00048 - (3.4 \times 10^{-5})$

Let us first convert 0.00048 to scientific notation

Steps to follow:

Move the decimal point in your number until there is only one non-zero digit to the left of the decimal point. The resulting decimal number is a.

Count how many places you moved the decimal point. This number is b.

If you moved the decimal to the left b is positive.

If you moved the decimal to the right b is negative.

If you did not need to move the decimal b = 0.

Write your scientific notation number as a x 10^b and read it as "a times 10 to the power of b."

Remove trailing 0's only if they were originally to the left of the decimal point.

$0.00048 = 4.8 \times 10^{-4}$

So the given expression becomes,

$\rightarrow 4.8 \times 10^{-4} - 3.4 \times 10^{-5}$

Let us make the exponent of second term as -4

$\rightarrow 4.8 \times 10^{-4} - 0.34 \times 10^{-4}$

Take $10^{-4}$ as common term,

$\rightarrow 10^{-4} (4.8 - 0.34)$

$\rightarrow 4.46 \times 10^{-4}$

Thus option C is correct

7 0

3 years ago

Use the definition of a Taylor series to find the first four nonzero terms of the series for f(x) centered at the given value of

Black_prince [1.1K]

Answer:

The first four nonzero terms of the Taylor series of $\frac{7}{x + 1}$ around $a=2$ are:

$f\left(x\right)\approx P\left(x\right) = \frac{7}{3}- \frac{7}{9}\left(x-2\right)+\frac{7}{27}\left(x-2\right)^{2}- \frac{7}{81}\left(x-2\right)^{3}+\frac{7}{243}\left(x-2\right)^{4}$

Step-by-step explanation:

The Taylor series of the function f at a (or about a or centered at a) is given by

$f\left(x\right)=\sum\limits_{k=0}^{\infty}\frac{f^{(k)}\left(a\right)}{k!}\left(x-a\right)^k$

To find the first four nonzero terms of the Taylor series of $\frac{7}{x + 1}$ around $a=2$ you must:

In our case,

$f\left(x\right) \approx P\left(x\right) = \sum\limits_{k=0}^{n}\frac{f^{(k)}\left(a\right)}{k!}\left(x-a\right)^k=\sum\limits_{k=0}^{4}\frac{f^{(k)}\left(a\right)}{k!}\left(x-a\right)^k$

So, what we need to do to get the desired polynomial is to calculate the derivatives, evaluate them at the given point, and plug the results into the given formula.

$f^{(0)}\left(x\right)=f\left(x\right)=\frac{7}{x + 1}$

Evaluate the function at the point: $f\left(2\right)=\frac{7}{3}$

$f^{(1)}\left(x\right)=\left(f^{(0)}\left(x\right)\right)^{\prime}=\left(\frac{7}{x + 1}\right)^{\prime}=- \frac{7}{\left(x + 1\right)^{2}}$

Evaluate the function at the point: $\left(f\left(2\right)\right)^{\prime }=- \frac{7}{9}$

$f^{(2)}\left(x\right)=\left(f^{(1)}\left(x\right)\right)^{\prime}=\left(- \frac{7}{\left(x + 1\right)^{2}}\right)^{\prime}=\frac{14}{\left(x + 1\right)^{3}}$

Evaluate the function at the point: $\left(f\left(2\right)\right)^{\prime \prime }=\frac{14}{27}$

$f^{(3)}\left(x\right)=\left(f^{(2)}\left(x\right)\right)^{\prime}=\left(\frac{14}{\left(x + 1\right)^{3}}\right)^{\prime}=- \frac{42}{\left(x + 1\right)^{4}}$

Evaluate the function at the point: $\left(f\left(2\right)\right)^{\prime \prime \prime }=- \frac{14}{27}$

$f^{(4)}\left(x\right)=\left(f^{(3)}\left(x\right)\right)^{\prime}=\left(- \frac{42}{\left(x + 1\right)^{4}}\right)^{\prime}=\frac{168}{\left(x + 1\right)^{5}}$

Evaluate the function at the point: $\left(f\left(2\right)\right)^{\prime \prime \prime \prime }=\frac{56}{81}$

Apply the Taylor series definition:

$f\left(x\right)\approx\frac{\frac{7}{3}}{0!}\left(x-\left(2\right)\right)^{0}+\frac{- \frac{7}{9}}{1!}\left(x-\left(2\right)\right)^{1}+\frac{\frac{14}{27}}{2!}\left(x-\left(2\right)\right)^{2}+\frac{- \frac{14}{27}}{3!}\left(x-\left(2\right)\right)^{3}+\frac{\frac{56}{81}}{4!}\left(x-\left(2\right)\right)^{4}$

The first four nonzero terms of the Taylor series of $\frac{7}{x + 1}$ around $a=2$ are:

$f\left(x\right)\approx P\left(x\right) = \frac{7}{3}- \frac{7}{9}\left(x-2\right)+\frac{7}{27}\left(x-2\right)^{2}- \frac{7}{81}\left(x-2\right)^{3}+\frac{7}{243}\left(x-2\right)^{4}$

8 0

3 years ago

Can someone please help me

77julia77 [94]

Symbolab has a calculator that solves problems like that and works it out for you, to help understand it better :)

5 0

3 years ago