Graph the line whose x-intercept is 8 and whose y-intercept is -9.

How much space is taken up by a book that is 12 inches long, 10 inches wide, and<br> 1 inch tall?

lara [203]

120 cause 12x10x1 equal ONE TWENTYYYY YEAHHAHAHAHAHAHAHHHHHAAHAHHHHH

3 0

2 years ago

Read 2 more answers

Find the measure of the missing angles! Pls help thank you

Inga [223]

Answer:

a=61°

Step-by-step explanation:

a= 180° - 90° - 29°

a = 61°

3 0

2 years ago

Is algebra.

Whitepunk [10]

Answer: the answer is d

Step-by-step explanation:

6 0

2 years ago

what is the equation of the line that is parallel to the given line and passes through the point (2, 3) ? a. x + 2y = 4 b. x + 2

aleksandr82 [10.1K]

Answer:

see explanations

Step-by-step explanation:

The given blue line has a slope of m = -1/2.

The line parallel to the given line passing through point (x0,y0)=(2,3) is given by the point-slope form:

(y-y0)=m(x-x0)

substitute values

(y-3) = (-1/2)(x-2)

Expand and transpose

y = (-1/2)x + 1 + 3

y = (-1/2)x + 4 ....................(1)

We choose the second equation b. x+2y=8 and convert to slope-intercept form:

2y=-x+8

y = (-1/2)x + 4, which is exactly equation (1)

So

b. x+2y=8 is the correct answer.

3 0

2 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

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

which converges everywhere. Then by substitution,

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

which also converges everywhere (and we can confirm this via the ratio test, for instance).

a. Differentiating the Taylor series gives

$f'(x)=\displaystyle4\sum_{n=1}^\infty(-1)^n\frac{nx^{4n-1}}{(2n)!}$

(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,

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

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

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