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Licemer1 [7]
3 years ago
12

What is 692,004 number in the other forms standard form?

Mathematics
2 answers:
Studentka2010 [4]3 years ago
6 0
Am presuming you mean in standard form:

692 004 =  6.92004 * 10⁵
ollegr [7]3 years ago
5 0

Answer:

The other standard form is 692004=692.004\times 10^3          

Step-by-step explanation:

Given : Number 692,004.

To find : What is number in the other forms standard form?

Solution :

Standard form is a way of writing down a number in either very large or very small numbers easily.

We have given the number 692,004

Writing in smaller number by multiplying and dividing by 1000.

The number became, 692004\times \frac{1000}{1000}=692.004\times 10^3

The other standard form is 692004=692.004\times 10^3

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Answer:

(0,-4) (5,0)

Step-by-step explanation:

4x-5y=20

First, we must put the equation into slope intercept form: y= mx+b

5y=4x-20=y=4/5x-4

The y intercept, by definition, is b. In this scenario, b is -4.

One ordered pair that works is (0,-4).

Also, we can find the ordered pair for the x intercept. By definition, the x intercept is the value of x when y equals zero. hence, we substitute 0 for y and then we can find an ordered pair that works and solve for x!:

  1. 0=4/5(x)-4= 4=4/5x
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Therefore, (5,0) works as well!

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3 years ago
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Is it d=8/3????......
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If y= -8 when x= -2 whats x when y=32
san4es73 [151]

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Find a compact form for generating functions of the sequence 1, 8,27,... , k^3
pantera1 [17]

This sequence has generating function

F(x)=\displaystyle\sum_{k\ge0}k^3x^k

(if we include k=0 for a moment)

Recall that for |x|, we have

\displaystyle\frac1{1-x}=\sum_{k\ge0}x^k

Take the derivative to get

\displaystyle\frac1{(1-x)^2}=\sum_{k\ge0}kx^{k-1}=\frac1x\sum_{k\ge0}kx^k

\implies\dfrac x{(1-x)^2}=\displaystyle\sum_{k\ge0}kx^k

Take the derivative again:

\displaystyle\frac{(1-x)^2+2x(1-x)}{(1-x)^4}=\sum_{k\ge0}k^2x^{k-1}=\frac1x\sum_{k\ge0}k^2x^k

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Take the derivative one more time:

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\implies\displaystyle\frac{x+4x^3+x^3}{(1-x)^4}=\sum_{k\ge0}k^3x^k

so we have

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