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FromTheMoon [43]
3 years ago
12

one section of an auditorium has 12 rows of seats each row has 13 seats what is the total number of seats in that section

Mathematics
2 answers:
Svet_ta [14]3 years ago
8 0
12 x 13 =156
There is 156 seats in that section
max2010maxim [7]3 years ago
5 0
The answer is 156 hope this helpes
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A six-sided number cube is labeled with the numbers 1-6, one number on each face. Each number is used exactly once.
I am Lyosha [343]

Answer:

36

Step-by-step explanation:

This question can be solved using fundamental counting principle which means to say that if there are "m" ways  of doing one thing and there are "n" ways of doing other thing then there are total "m*n" ways of doing other things.

For example: if there are 4 type of pens and 3 type of notebook then there are "4*3" ways to write on notebook using  the pens.

_____________________________________________________

Using the above mentioned principle we have

 six sided cube with number 1 to 6  written on it hence in first roll there are six possible outcome(1 or 2 or 3 or 4 or 5 or 6) similarly there are 6 possible outcome in other roll as well.

If the combination of outcome is to be deduced then there are 6*6 ways of outcome based on  fundamental counting principle.

Hence there are 6*6 = 36 ways of possible outcomes.

_________________________________________________________

8 0
3 years ago
WILL GIVE BRAINLIEST
tankabanditka [31]

Answer:

2

Step-by-step explanation:

It said it was 2 in edg. Have a great day.

3 0
3 years ago
Sketch over the interval [0, 2π] the graph of: y = sin(x + π)<br>​
LuckyWell [14K]

Answer:

solve the inequality /2x-1/≥3

7 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
Please help with a math problem :/ Please show steps...<br><br> x^2 -13x = -36
Marianna [84]
X^2 -  13 x = -36 
add 36 to both sides

x^2 - 13x +36 = 0
figure out the factor
(x-4)(x-9) = 0
so (x-4) or (x-9) equals 0
so solution is
x-4=0 add 4 to both sides and get x = 4
x-9=0 add 9 to both sides and get x=9
x={ 4, 9}
4 0
3 years ago
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