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3 years ago

If mary can bake more cakes in one day than sarah can bake in one day, then _________

Mathematics

1 answer:

Debora [2.8K]3 years ago

4 0

I'm not sure, but it's probably that Mary bakes faster than Sarah?

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2 Points Which of the following is(are) the solution(s) to . x+6 = 2x - 1?

seropon [69]

Answer:

x = 7

Step by step explanation:

minus x on both sides

6 = x -1

add 1 on both sides

x = 7

7 0

3 years ago

U^2+u-72/3u-15 need to factor completely

lesya [120]

$U^2+u-72/3u-15$

$u^2+u-24u-15
$

$u^2-23u-15
$

3 0

3 years ago

How do you know if a graph you are looking at shows a periodic function? How do you determine the period of the graph?

mylen [45]

Answer:

A function f(x) is said to be periodic, if there exists a positive real number T such that f(x+T) = f(x).

You can also just say: A periodic function is one that repeats itself in regular intervals.

Step-by-step explanation:

The smallest value of T is called the period of the function.

Note: If the value of T is independent of x then f(x) is periodic, and if T is dependent, then f(x) is non-periodic.

For example, here's the graph of sin x. [REFER TO PICTURE BELOW]

Sin x is a periodic function with period 2π because sin(x+2π)=sinx

Other examples of periodic functions are all trigonometric ratios, fractional x (Denoted by {x} which has period 1) and others.

In order to determine the period of the determined graph however, just know that the period of the sine curve is the length of one cycle of the curve. The natural period of the sine curve is 2π. So, a coefficient of b=1 is equivalent to a period of 2π. To get the period of the sine curve for any coefficient b, just divide 2π by the coefficient b to get the new period of the curve.

Hopefully this helped a bit.

6 0

3 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

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