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erica [24]
3 years ago
11

20. Which problems will have two decimal places in the product?Mark all that apply.531 x 107.4 x 10(A) 5 X 0.89E 3.2 x 43(D) 6.1

x 3

Mathematics
1 answer:
Schach [20]3 years ago
7 0
A and E When you multiply any equation with a decimal you add that many decimal places in the end. Ex: If you multiply 4.12×5.8 there will be 3 decimal places in the product because 4.12 had 2 decimal places and 5.8 had 1 decimal place and 2+1=3. The only exception is when you multiply a decimal with a number that ends with a 0 and is not a decimal like 10, 20, 30, 100, 1000, etc. If this is the case then put the decimal where it would have been and then move it right the same number as you have 0s. Ex: 1000× 5.82 would have been 5.82000*, but becomes 5,820.00*, which really is 5,820*. *continuous 0s after a decimal is unnecessary and will probably make you lose points on a test, but I was trying to prove a point
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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
3 years ago
How do you cauculate net change ​
Travka [436]

Answer:

You can calculate net change by subtracting the current day's closing price for an asset from the closing price of the previous day!

Step-by-step explanation:

(What net change is) The net change theorem states that when a quantity changes, the final value equals the initial value plus the integral of the rate of change. Net change can be a positive number, a negative number, or zero.

NOT MY ANSWER OR WHATEVER I GOT IT FROM G0OGLE

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asambeis [7]

Answer:

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Step-by-step explanation:

6 0
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The length is 10
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