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cupoosta [38]
2 years ago
9

Don’t solve just write the equation!!!

Mathematics
1 answer:
erastovalidia [21]2 years ago
8 0

Answer:

2(x - 4) = 16

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Which of the following is the inverse of the function {(1,2),(3,4),(6,8)}
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C.<span>{(2,1),(4,3),(8,6)}</span>
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Given m = (
DIA [1.3K]

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i need to know the question to answer ir

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I need help on thi and the first person who answer correctly gets a BRANLIST​ and show work​
const2013 [10]

Answer:

w<6 THE SMALL LINE UNDER THE LESS THAN SIGN IS STILL THERE I JUST CANNOT ADD IT

Step-by-step explanation:

first solve the equation to the right.

-3(2w+1) = -6w-3

now solve the whole equation

-33-w< -6w-3

-w+6w < -3+33

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divide the 5 from both sides to simplify

5/5w < 30/5

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THE ANSWER IS w < 6

3 0
2 years ago
I write out all the whole numbers, starting from 1. If I wrote 1994 digits altogether, what was the last complete number I wrote
Nataly_w [17]

Answer:

You wrote the numbers from 1 to 9: 9 digits.

You wrote the numbers from 10 to 99: 90*2 = 180 digits.

You have written 189 digits.

Then you wrote another 1994 - 189 = 1805

All the numbers from 100 to 999 have 3 digits

So you wrote 1805/3 = 601 2/3

This means 601 full numbers and 2 digits from the 602. number.

So 100 + 601 - 1 = 700 was the last number you wrote.

Do you agree?


3 0
3 years ago
A) How many ways can 2 integers from 1,2,...,100 be selected
Anna007 [38]

Answer with explanation:

→Number of Integers from 1 to 100

                                            =100(50 Odd +50 Even)

→50 Even =2,4,6,8,10,12,14,16,...............................100

→50 Odd=1,3,,5,7,9,..................................99.

→Sum of Two even integers is even.

→Sum of two odd Integers is odd.

→Sum of an Odd and even Integer is Odd.

(a)

Number of ways of Selecting 2 integers from 50 Integers ,so that their sum is even,

   =Selecting 2 Even integers from 50 Even Integers , and Selecting 2 Odd integers from 50 Odd integers ,as Order of arrangement is not Important, ,

        =_{2}^{50}\textrm{C}+_{2}^{50}\textrm{C}\\\\=\frac{50!}{(50-2)!(2!)}+\frac{50!}{(50-2)!(2!)}\\\\=\frac{50!}{48!\times 2!}+\frac{50!}{48!\times 2!}\\\\=\frac{50 \times 49}{2}+\frac{50 \times 49}{2}\\\\=1225+1225\\\\=2450

=4900 ways

(b)

Number of ways of Selecting 2 integers from 100 Integers ,so that their sum is Odd,

   =Selecting 1 even integer from 50 Integers, and 1 Odd integer from 50 Odd integers, as Order of arrangement is not Important,

        =_{1}^{50}\textrm{C}\times _{1}^{50}\textrm{C}\\\\=\frac{50!}{(50-1)!(1!)} \times \frac{50!}{(50-1)!(1!)}\\\\=\frac{50!}{49!\times 1!}\times \frac{50!}{49!\times 1!}\\\\=50\times 50\\\\=2500

=2500 ways

7 0
2 years ago
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