Which equation shows a slope of 3 and a y-intercept of (0,7) ?

Mathematics

2 answers:

asambeis [7]3 years ago

5 0

Answer:

y=3x+7

Step-by-step explanation:

The equation of a line in slope-intercept form is:

y=mx+b

where m is the slope and b is the y-intercept.

We know the slope is 3, so we can substitute 3 in for m.

y=3x+b

We also know the y-intercept is (0,7). When writing the equation of a line in this form, we can ignore the x-coordinate of 0. Therefore, the y-intercept is also just 7. Substitute 7 in for b.

y=3x+7

The equation of a line with a slope of 3 and a y-intercept of (0,7) is y=3x+7

12345 [234]3 years ago

3 0

Answer: y = 3x + 7

Step-by-step explanation:

Since we are given the slope and y intercept we could write the equation in slope intercept form as y=mx +b .Only m and b are needed to write the equation.M is the slope and B is the y intercept.

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If f (n)(0) = (n + 1)! for n = 0, 1, 2, , find the taylor series at a=0 for f.

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Given that $f^{(n)}(0)=(n+1)!$ , we have for $f(x)$ the Taylor series expansion about 0 as

$f(x)=\displaystyle\sum_{n=0}^\infty\frac{(n+1)!}{n!}x^n=\sum_{n=0}^\infty(n+1)x^n$

Replace $n+1$ with $n$ , so that the series is equivalent to

$f(x)=\displaystyle\sum_{n=1}^\infty nx^{n-1}$

and notice that

$\displaystyle\frac{\mathrm d}{\mathrm dx}\sum_{n=0}^\infty x^n=\sum_{n=1}^\infty nx^{n-1}$

Recall that for $|x|$ , we have

$\displaystyle\sum_{n=0}^\infty x^n=\frac1{1-x}$

which means

$f(x)=\displaystyle\sum_{n=1}^\infty nx^{n-1}=\frac{\mathrm d}{\mathrm dx}\frac1{1-x}$
$\implies f(x)=\dfrac1{(1-x)^2}$

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