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baherus [9]
3 years ago
8

Find the slope of a line given the following points

Mathematics
1 answer:
nasty-shy [4]3 years ago
6 0

Answer:

  -6/5

Step-by-step explanation:

The slope is computed from ...

  slope = (change in y)/(change in x)

  = (-2 -4)/(3 -(-2)) = -6/5

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If U=(1,2,3,4,5,6,7,8,9) <br><br> A=(1,2,3) is A a subset of U and is U a superset of A.
barxatty [35]

Answer:

yes A is a SUBSET of U and U is a SUPERSET of A.

Explanation: use curly braces for the elements.

A is a subset of U since U has the elements {1,2,3,4,5,6,7,8,9}, and A has the elements {1,2,3}

U contains the elements listed in set A, so a is a SUBSET of U

7 0
3 years ago
Read 2 more answers
Which of the following is the estimated sum of fifty-five, sixty, and sixty-five using compatible numbers?
SVEN [57.7K]

C. 190

Round each number

55→60.

60→60

65→70

Then add all of your rounded numbers.

60+60+70=190

3 0
3 years ago
Find the value of in the triangle shown below<br><br>9, 7, and x around the triangle <br>​
borishaifa [10]

Answer:

Step-by-step explanation: 56

3 0
3 years ago
Plzzz help....I am timed...There will be a picture provided....You will get 30 points!!! MATHH
xxMikexx [17]

Answer:

It should be the last one, She made a sign error when multiplying.

3 0
3 years ago
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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
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