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nirvana33 [79]
3 years ago
9

Evaluate the expression 5* – 31 for = 2. Help pls!

Mathematics
1 answer:
salantis [7]3 years ago
5 0

Answer:

the answer is 16

Step-by-step explanation:

5^2-3^2

25-9

16

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1 5 y + 8 − 4 5 = 9 10 y − 9 20
emmainna [20.7K]

Answer:

y ≈  .98   i hope this helps :)

Step-by-step explanation:

given equation   15y + 8 - 45 = 910y - 920

simplify the equation   15y - 37 = 910y - 920

subtract 15y from both sides   -37 = 895y - 920

add 920 to both sides   883 = 895y

divide both sides by 895   .98 ≈ y

y ≈  .98

0.98659217877

7 0
3 years ago
Can someone help me with thos
Harman [31]

Answer: 65

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
What is the probability of getting a 6 each time if a die is rolled 4 times?
aleksley [76]
Kinda low, you would need luck to be on your side!
4 0
4 years ago
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What percent of 72 is 48 with step by step process
Sedaia [141]

Answer:

67%

Step-by-step explanation:

There are a few methods to finding the percent of a number.  You can use proportions:

\frac{percent}{100}=\frac{48}{72}

The 'percent' is the percent of the fraction that you are looking for, or 48/72.

Cross-multiply:  72x = (48)(100) or 72x = 4800

Divide both sides by 72:  72x/72 = 4800/72

Solve for 'x':  x = 66.6666...

Round 66.666.... to the nearest percent: 67%

8 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
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

#SPJ4

3 0
2 years ago
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