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lubasha [3.4K]
3 years ago
12

What is 0.01 as a fraction

Mathematics
2 answers:
AnnyKZ [126]3 years ago
8 0

Answer:

hey! this is simple! .01 is 1 out of 100.

Step-by-step explanation:

so the answe is, 1/100

USPshnik [31]3 years ago
7 0
1 over 10

Explainedddddd-
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How are the two functions f(x) = 0.7(6)x and g(x) = 0.7(6)–x related to each other?
Harlamova29_29 [7]
The answer
f(x) = 0.7(6)x = <span>f(x) = 0.7(6)^x, and  </span><span>g(x) = 0.7(6)–x= </span>g(x) = 0.7(6)^-x=1/<span>0.7(6)^x
so </span>
g(x) =1/<span>0.7(6)^x=1 /</span><span><span>f(x)

</span> the relationship between f and g are </span>g(x) =1 /<span>f(x) or </span><span>g(x) . <span>f(x) = 1</span> </span>






7 0
3 years ago
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How many rational numbers have square roots between 10 and 10.1?​
Ymorist [56]

Answer:

0.1

Step-by-step explanation:

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Identify the usual level of measurement for each of the following A. year in school B. IQ scores C. life expectancy D. fatigue E
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Step-by-step explanation:

4 0
2 years ago
Calculate the sum of the multiples of 4 from 0 to 1000
allochka39001 [22]

Answer:

sum is 125,500

sum in summation notation is \sum\limits_{n=0}^n a+nd= (2a+(n-1)d)n/2

Step-by-step explanation:

This problem can be solved using concept of arithmetic progression.

The sum of n term terms in arithmetic progression is given by

sum = (2a+(n-1)d)n/2

where

a is the first term

d is the common difference of arithmetic progression

_____________________________________________________

in the problem

series is multiple of 4 starting from 4 ending at 1000

so series will look like

series: 0,4,8,12,16..................1000

a is first term so

here a is 0

lets find d the common difference

common difference is given by nth term - (n-1)th term

lets take nth term as 8

so (n-1)th term = 4

Thus,

d = 8-4 = 4

d  can also be seen 4 intuitively as series is multiple of four.

_____________________________________________

let calculate value of n

we have last term as 1000

Nth term can be described

Nth term = 0+(n-1)d

1000 =   (n-1)4

=> 1000 = 4n -4

=> 1000 + 4= 4n

=> n = 1004/4 = 251

_____________________________________

now we have

n = 1000

a = 0

d = 4

so we can calculate sum of the series by using formula given above

sum = (2a+(n-1)d)n/2

       = (2*0 + (251-1)4)251/2

       = (250*4)251/2

     = 1000*251/2 = 500*251 = 125,500

Thus, sum is 125,500

sum in summation notation is \sum\limits_{n=0}^n a+nd= (2a+(n-1)d)n/2

3 0
2 years ago
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