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Svetllana [295]
2 years ago
14

What is an equation of the line that passes through the points (0,7) and (5,4? Put your answer in fully reduced form.

Mathematics
1 answer:
gregori [183]2 years ago
4 0

Answer <u>(assuming it can be in slope-intercept format)</u>:

y = -\frac{3}{5} x+7  

Step-by-step explanation:

When knowing the y-intercept of a line and its slope, we can write an equation representing it in slope-intercept form, or y = mx + b.

1) First, find the slope of the equation. Use the slope formula, m = \frac{y_2-y_1}{x_2-x_1}, to find the slope. Substitute the x and y values of the given points into the formula and simplify:

m = \frac{(4)-(7)}{(5)-(0)} \\m = \frac{4-7}{5-0} \\m = \frac{-3}{5}

Thus, the slope is -\frac{3}{5}.

2) Usually, we would have to use one of the given points and the slope to put the equation in point-slope form. However, notice that the point (0,7) has an x-value of 0. All points on the y-axis have an x-value of 0, thus (0,7) must be the y-intercept of the line. Now that we know the slope of the line and its y-intercept, we can already write the equation in slope-intercept format, represented by the equation  y = mx + b. Substitute m and b for real values.

Since m represents the slope, substitute -\frac{3}{5}  in its place in the equation. Since b represents the y-intercept, substitute 7 in its place. This gives the following equation and answer:

y = -\frac{3}{5} x+7

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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:

<u>These are the steps to find the Taylor series for the function</u> sin^2(3 x)

  1. 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)!}

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

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}

<u>To find the interval on which the series converges you need to use the Ratio Test that says</u>

For the power series centered at x=a

P(x)=C_{0}+C_{1}(x-a)+C_{2}(x-a)^{2}+...+ C_{n}(x-a)^{n}+...,

suppose that \lim_{n \to \infty} |\frac{C_{n}}{C_{n+1}}| = R.. Then

  • If R=\infty, the the series converges for all x
  • If 0 then the series converges for all |x-a|
  • If R=0, the the series converges only for x=a

So we need to evaluate this limit:

\lim_{n \to \infty} |\frac{\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}}{\frac{-9^{n+1}2^{2*(n+1)-1}x^{2*(n+1)}}{(2*(2n+1))!}} |

Simplifying we have:

\lim_{n \to \infty} |-\frac{(n+1)(2n+1)}{18x^{2} } |

Next we need to evaluate the limit

\lim_{n \to \infty} |-\frac{(n+1)(2n+1)}{18x^{2} } |\\ \frac{1}{18x^{2} } \lim_{n \to \infty} |-(n+1)(2n+1)}|}

-(n+1)(2n+1) is negative when n -> ∞. Therefore |-(n+1)(2n+1)}|=2n^{2}+3n+1

You can use this infinity property \lim_{x \to \infty} (ax^{n}+...+bx+c) = \infty when a>0 and n is even. So

\lim_{n \to \infty} |-\frac{(n+1)(2n+1)}{18x^{2} } | \\ \frac{1}{18x^{2}} \lim_{n \to \infty} 2n^{2}+3n+1=\infty

Because this limit is ∞ the radius of converge is ∞ and the interval of converge is ( -\infty, \infty ).

6 0
3 years ago
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