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Shalnov [3]
3 years ago
12

What is 23 over 18 in simplest form

Mathematics
1 answer:
tatuchka [14]3 years ago
6 0
\frac{23}{18} is already in simplest form because 23 is a prime number, which means that it is only divisible by 1 and itself (23). Hope this helps! :)
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Use the appropriate substitutions to write down the first four nonzero terms of the Maclaurin series for the binomial (1+3x)^(-1
PolarNik [594]

Answer:

First term=1

Second term=-x

Third term=2x^2

Fourth term =-\frac{28}{3!}x^3

Step-by-step explanation:

We are given that function

f(x)=(1+3x)^{-1/3}

We have to find the  first four non zero terms of the Maclaurin series for the binomial.

Maclaurin series of function f(x) is given by

f(x)=f(0)+f'(0)x+\frac{1}{2!}f''(0)x^2+\frac{1}{3!}f'''(0)x^3+....

f(0)=(1+3x)^{\frac{-1}{3}}=1

f'(x)=-\frac{1}{3}(1+3x)^{-\frac{4}{3}}(3)=-(1+3x)^{-\frac{4}{3}}

f'(0)=-1

f''(x)=\frac{4}{3}\times 3 (1+3x)^{-\frac{7}{3}}

f''(0)=4

f'''(x)=-4\times \frac{7}{3}\times 3(1+3x)^{-\frac{10}{3}}

f'''(0)=-28

Substitute the values we get

(1+3x)^{-\frac{1}{3}}=1-x+\frac{4}{2!}x^2+\frac{-28}{3!}x^3+...

(1+3x)^{-\frac{1}{3}}=1-x+2x^2+\frac{-28}{3!}x^3+...

First term=1

Second term=-x

Third term=2x^2

Fourth term =-\frac{28}{3!}x^3

5 0
3 years ago
Find the 66th term of the arithmetic sequence -28,-45, -62,-
Alexeev081 [22]

Answer:

<u>-1133</u>

Step-by-step explanation:

<u>Formula for nth term</u>

  • aₙ = a + (n - 1)d

<u>Finding d</u>

  • d = -45 - (-28)
  • d = -45 + 28
  • d = -17

<u>Solving for a₆₆</u>

  • a₆₆ = -28 + 65(-17)
  • a₆₆ = -28 - 1105
  • a₆₆ = <u>-1133</u>
6 0
2 years ago
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aleksandr82 [10.1K]
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(0,6)
(4,9)

(3,6)
(2,3)

These points can be substituted into the systems of equation in the choices and inspect which equations satisfy the value of the points. Doing this, the answer is
3x - 4y = -24
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I think it’s b ????
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