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harina [27]
3 years ago
11

Help please!

Mathematics
1 answer:
Alina [70]3 years ago
7 0

Step-by-step explanation:

Given

f(x) = 8x - 9

Then

For a.

f(3y - 1 ) = 8(3y - 1 ) - 9

= 24y - 8 - 9

= 24y - 17

Now for b.

f(x) = - 23

or, 8x - 9 = - 23

8x = - 23 + 9

8x = - 14

x = - 14 / 8

x = - 7 / 4

Hope it will help :)

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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
How would you find the surface area of the figure represented by the given net? (Note: The area of each circular base is the sam
ioda

Answer:

Add 1,105.28 + 200.96 + 200.96.

Step-by-step explanation:

we know that

The surface area of the cylinder is equal to

SA=2B+LA

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LA is the lateral area

In this problem we have

B=200.96\ mm^{2}

LA=1,105.28\ mm^{2}

substitute

SA=2(200.96)+1,105.28

SA=1,105.28+200.96+200.96

SA=1,507.2\ mm^{2}

7 0
4 years ago
Read 2 more answers
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mote1985 [20]
Her ham would cost $3.95
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The weight of an object on moon is 1/6 of its weight on Earth. If weight of an object on Earth is 18/5 kg, what would be its wei
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Answer:

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The weight of an object on Moon is 6 times smaller than same on Earth.

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