Simplify each expression (s-v)2

A light bulb consumes 600 watt-hours per day. How long does it take to consume 2250 watt-hours?

Alexus [3.1K]

If a light bulb consumes 600 per day, then you need to divide 2250 by 600. So it would be around 3.75 days

3 0

4 years ago

What is the final amount if 893 is decreased by 2% followed by a further 7% decrease?

Ahat [919]

Answer 62 because if you work it out it wi give it

6 0

3 years ago

HELP ASAP ILL GIVE You 5 stars<br> What is the solution to this inequality?<br> 7x + 25<=11

JulsSmile [24]

Answer:

Step-by-step explanation:

smplifying

7x + 11 = 25

Reorder the terms:

11 + 7x = 25

Solving

11 + 7x = 25

Solving for variable 'x'.

Move all terms containing x to the left, all other terms to the right.

Add '-11' to each side of the equation.

11 + -11 + 7x = 25 + -11

Combine like terms: 11 + -11 = 0

0 + 7x = 25 + -11

7x = 25 + -11

Combine like terms: 25 + -11 = 14

7x = 14

Divide each side by '7'.

x = 2

Simplifying

x = 2

3 0

3 years ago

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Cara has another candle that is 15 cm tall. How fast must it burn in order to also be 6 cm tall after 4 hours?

egoroff_w [7]

Answer:

It has to burn at a rate of 2.25 cm per hour.

Step-by-step explanation:

4 0

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

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

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

4 years ago