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3 years ago

Please helpppp. It's due in an hour

Mathematics

1 answer:

TEA [102]3 years ago

4 0

Answer:

sowwy

Step-by-step explanation:

i can't help WITH MY SMALL BRAIN

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PLEASE ANSWER ASAP! MUST BE CORRECT! MORE POINTS GIVEN!!!

Tasya [4]

These are the two rules for when a and b are positive numbers.
a + b = b + a
a - b ≠ b -a
a - b = -b + a

For example:

5.71 + 2.84 = 2.84 + 5.71
8.55 = 8.55

5.71 - 2.84 ≠ 2.84 - 5.71
2.87 ≠ -2.87

5.71 - 2.84 = -2.84 + 5.71
2.87 = 2.87

These are the rules for when a and b are negative numbers.
a + b = b + a
a - b = b + a

For example,

-6.2 + (-3.96) = -3.96 + (-6.2)
-6.2 - 3.96 = -3.96 - 6.2
-10.16 = -10.16

-6.2 - (3.96) = -3.96 + (-6.2)
-10.16 = -10.16

Also, if a is a positive number, while b is a negative number, we see these rules:
a + b = a - b
a - b = a + b

For example,

5.71 + (-6.2) = 5.71 - 6.2
-0.49 = -0.49

5.71 - (-6.2) = 5.71 + 6.2
11.91 = 11.91

Also, if a is a negative number while b is a positive number, then these rules will apply:
a + b = b - a
a - b = -b - a

For example,

-3.96 + 2.84 = 2.84 - 3.96
-1.12 = -1.12

-3.96 - 2.84 = -2.84 - 3.96
-6.8 = -6.8

I hope this helps! :)

4 0

2 years ago

Use this list of Basic Taylor Series and the identity sin2θ= 1 2 (1−cos(2θ)) to find the Taylor Series for f(x) = sin2(3x) based

notsponge [240]

Answer:

The Taylor series for $sin^2(3 x) = - \sum_{n=1}^{\infty} \frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}$ , the first three non-zero terms are $9x^{2} -27x^{4}+\frac{162}{5}x^{6}$ and the interval of convergence is $( -\infty, \infty )$

Step-by-step explanation:

These are the steps to find the Taylor series for the function $sin^2(3 x)$

Use the trigonometric identity:

$sin^{2}(x)=\frac{1}{2}*(1-cos(2x))\\ sin^{2}(3x)=\frac{1}{2}*(1-cos(2(3x)))\\ sin^{2}(3x)=\frac{1}{2}*(1-cos(6x))$

2. The Taylor series of $cos(x)$

$cos(y) = \sum_{n=0}^{\infty}\frac{-1^{n}y^{2n}}{(2n)!}$

Substituting y=6x we have:

$cos(6x) = \sum_{n=0}^{\infty}\frac{-1^{n}6^{2n}x^{2n}}{(2n)!}$

3. Find the Taylor series for $sin^2(3x)$

$sin^{2}(3x)=\frac{1}{2}*(1-cos(6x))$ (1)

$cos(6x) = \sum_{n=0}^{\infty}\frac{-1^{n}6^{2n}x^{2n}}{(2n)!}$ (2)

Substituting (2) in (1) we have:

$\frac{1}{2} (1-\sum_{n=0}^{\infty}\frac{-1^{n}6^{2n}x^{2n}}{(2n)!})\\ \frac{1}{2}-\frac{1}{2} \sum_{n=0}^{\infty}\frac{-1^{n}6^{2n}x^{2n}}{(2n)!}$

Bring the factor $\frac{1}{2}$ inside the sum

$\frac{6^{2n}}{2}=9^{n}2^{2n-1} \\ (-1^{n})(9^{n})=(-9^{n} )$

$\frac{1}{2}-\sum_{n=0}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}$

Extract the term for n=0 from the sum:

$\frac{1}{2}-\sum_{n=0}^{0}\frac{-9^{0}2^{2*0-1}x^{2*0}}{(2*0)!}-\sum_{n=1}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}\\ \frac{1}{2} -\frac{1}{2} -\sum_{n=1}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}\\ 0-\sum_{n=1}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}\\ sin^{2}(3x)=-\sum_{n=1}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}$

To find the first three non-zero terms you need to replace n=3 into the sum

$sin^{2}(3x)=\sum_{n=1}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}\\ \sum_{n=1}^{3}\frac{-9^{3}2^{2*3-1}x^{2*3}}{(2*3)!} = 9x^{2} -27x^{4}+\frac{162}{5}x^{6}$