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Elden [556K]
3 years ago
14

What is...a increased by 25%

Mathematics
2 answers:
Gelneren [198K]3 years ago
7 0

Answer:

  1..25a

Step-by-step explanation:

a + 25%·a = a(1 + 0.25) = 1.25a

_____

<em>Comment on percentages</em>

The symbol "%" is a shorthand way to write "/100", so 25% = 25/100 = 0.25.

Of course, 100/100 = 100% = 1.

Adding 25% of something to 100% of it gives 125% of it, the original amount multiplied by 125/100 = 1.25.

BigorU [14]3 years ago
5 0

Answer:

1.25a

Step-by-step explanation:

a + 25%·a = a(1 + 0.25) = 1.25a

_____

Comment on percentages

The symbol "%" is a shorthand way to write "/100", so 25% = 25/100 = 0.25.

Of course, 100/100 = 100% = 1.

Adding 25% of something to 100% of it gives 125% of it, the original amount multiplied by 125/100 = 1.25.

Read more on Brainly.com - brainly.com/question/9011267#readmore

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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
Bond [772]

Taylor series is f(x) = ln2 + \sum_{n=1)^{\infty}(-1)^{n-1} \frac{(n-1)!}{n!(9)^{n}(x9)^{2}  }

To find the Taylor series for f(x) = ln(x) centering at 9, we need to observe the pattern for the first four derivatives of f(x). From there, we can create a general equation for f(n). Starting with f(x), we have

f(x) = ln(x)

f^{1}(x)= \frac{1}{x} \\f^{2}(x)= -\frac{1}{x^{2} }\\f^{3}(x)= -\frac{2}{x^{3} }\\f^{4}(x)= \frac{-6}{x^{4} }

.

.

.

Since we need to have it centered at 9, we must take the value of f(9), and so on.

f(9) = ln(9)

f^{1}(9)= \frac{1}{9} \\f^{2}(9)= -\frac{1}{9^{2} }\\f^{3}(x)= -\frac{1(2)}{9^{3} }\\f^{4}(x)= \frac{-1(2)(3)}{9^{4} }

.

.

.

Following the pattern, we can see that for f^{n}(x),

f^{n}(x)=(-1)^{n-1}\frac{1.2.3.4.5...........(n-1)}{9^{n} }  \\f^{n}(x)=(-1)^{n-1}\frac{(n-1)!}{9^{n}}

This applies for n ≥ 1, Expressing f(x) in summation, we have

\sum_{n=0}^{\infinite} \frac{f^{n}(9) }{n!} (x-9)^{2}

Combining ln2 with the rest of series, we have

f(x) = ln2 + \sum_{n=1)^{\infty}(-1)^{n-1} \frac{(n-1)!}{n!(9)^{n}(x9)^{2}  }

Taylor series is f(x) = ln2 + \sum_{n=1)^{\infty}(-1)^{n-1} \frac{(n-1)!}{n!(9)^{n}(x9)^{2}  }

Find out more information about taylor series here

brainly.com/question/13057266

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3 0
2 years ago
Round 5,836,197 to the nearest humdred
Olenka [21]
5,836,200 Is the answer :) Have a Great day!!!
4 0
3 years ago
Read 2 more answers
This is really confusing. Thanks for helping.
Mandarinka [93]
Count change of signs to tell positive roots
then replace x with -x and evaluate, then count change in sign for negative roots


initially
+, -, -, +, +, -
3 changes
3 or 1 positive roots


replacing x with -x
-,-,+,+,-,-
2 changes
2 or 0 negative roots



B is answer
5 0
3 years ago
Michael’s weight can be represented by the expression 72x^5. Al’s weight can be represented by the expression 9x^7.
podryga [215]

Answer:

a) \frac{72x^5}{9x^7}

b) =\frac{8}{x^2}

c), Yes;8x^2

Step-by-step explanation:

Michael's Weight: 72x^5

Al's weight: 9x^7

a) Ratio of Michael's weight to Al's weight: \frac{72x^5}{9x^7}

b) This ratio simplifies to:  \frac{8\times 9x^5}{9x^5\timesx^2}

                                         =\frac{8}{x^2}

c) Yes, If the exponent in each expression were negative, then we have:

Ratio of Michael's weight to Al's weight: \frac{72x^{-5}}{9x^{-7}}

This ratio simplifies to:  8x^{-5--7}=8x^2

The two ratios are not the same.

3 0
3 years ago
What is the reciprocal of 7 1/3?
Andrei [34K]
The reciprocal is -3/23.
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3 years ago
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