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3 years ago

Round 428,213 to the nearest ten thousand

Mathematics

1 answer:

givi [52]3 years ago

6 0

Answer:

Round 428,213 to the nearest ten thousand = 428,210

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How does algebra affect how people can learn from each other?

pantera1 [17]

Algebra is Faster And Better Than “Basic” Math

Just as multiplying two by twelve is faster than counting to 24 or adding 2 twelve times, algebra helps us solve problems more quickly and easily than we could otherwise. Algebra also opens up whole new areas of life problems, such as graphing curves that cannot be solved with only foundational math skills.

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3 years ago

If 4x + 8x + 12x + 16x is equal to 5 + 10 + 30 + 40 then what is the value of x

zloy xaker [14]

4x+8x+12x+16x=5+10+30+40

4x+8x+12x+16x= 85

40x=85

x= 85/40

x=2.125

7 0

3 years ago

(b-2)x= 8

s344n2d4d5 [400]

Answer:

Step-by-step explanation:

(2-2)x = 8

(0)x = 8

x = 8/0

no solution

6 0

3 years ago

Pls help me with this and an explanation

SCORPION-xisa [38]

Answer:

The relationship in this equation is linear, but not proportional. If it was proportional, then replacing x with nx should give ny as the answer. Example:

y = 3x

Replace x with 2x.

y' = 3(2x) = 2(3x) = 2y

y' = 2y

This is proportional. But if you replace x with 2x in

y = 3x + 5,

you will not get 2y for y'. This case is not proportiona

5 0

2 years ago

Read 2 more answers

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