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2 years ago

If an equation of a line is y - 5 = m(x-2) and (1,1) is a point on the graph, what is m?

Mathematics

2 answers:

arlik [135]2 years ago

7 0

Answer:

m=4

Step-by-step explanation:

given equation is

y-5=m(x-2)

put given point in above equation

1-5=m(1-2)

-4=m(-1)

-4=-m

m=4

BartSMP [9]2 years ago

3 0

Answer:

m = 4

Step-by-step explanation:

Substitute x = 1, y = 1 , that is(1,1) into the equation

1 - 5 = m(1 - 2)

- 4 = - m ( multiply both sides by - 1 )

m = 4

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3 years ago

Consider the function f(x)equalscosine left parenthesis x squared right parenthesis. a. Differentiate the Taylor series about 0

dybincka [34]

I suppose you mean

$f(x)=\cos(x^2)$

Recall that

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

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$\cos(x^2)=\displaystyle\sum_{n=0}^\infty(-1)^n\frac{(x^2)^{2n}}{(2n)!}=\sum_{n=0}^\infty(-1)^n\frac{x^{4n}}{(2n)!}$

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(starting at $n=1$ because the summand is 0 when $n=0$ )

b. Naturally, the differentiated series represents

$f'(x)=-2x\sin(x^2)$

To see this, recalling the series for $\sin x$ , we know

$\sin(x^2)=\displaystyle\sum_{n=0}^\infty(-1)^{n-1}\frac{x^{4n+2}}{(2n+1)!}$

Multiplying by $-2x$ gives

$-x\sin(x^2)=\displaystyle2x\sum_{n=0}^\infty(-1)^n\frac{x^{4n}}{(2n+1)!}$

and from here,

$-2x\sin(x^2)=\displaystyle 2x\sum_{n=0}^\infty(-1)^n\frac{2nx^{4n}}{(2n)(2n+1)!}$

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c. This series also converges everywhere. By the ratio test, the series converges if

$\displaystyle\lim_{n\to\infty}\left|\frac{(-1)^{n+1}\frac{(n+1)x^{4(n+1)}}{(2(n+1))!}}{(-1)^n\frac{nx^{4n}}{(2n)!}}\right|=|x|\lim_{n\to\infty}\frac{\frac{n+1}{(2n+2)!}}{\frac n{(2n)!}}=|x|\lim_{n\to\infty}\frac{n+1}{n(2n+2)(2n+1)}$

The limit is 0, so any choice of $x$ satisfies the convergence condition.

3 0

3 years ago

Please help me thanks

marissa [1.9K]

Answer:

The constant rate of change is 1.5

The point that intersects the y axis is (0,0)

Step-by-step explanation:

Constant rate of change: y2-y1/x2-x1

(1,1.5) (2,3)

3-1.5=1.5

2-1=1

1.5/1=1.5

Constant rate of change is 1.5

Hope this helps!

8 0

2 years ago