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ivann1987 [24]
2 years ago
6

(x=1)^2-4x=(x-1)^2 simplify and answer

Mathematics
1 answer:
Dennis_Churaev [7]2 years ago
5 0

Step-by-step explanation:

Firstly, there's a typing error at (x=1)². I assume it is (x-1)². If my assumption is correct,then:

x²-2x+1-4x=x²-2x+1

x²-6x+1-x²+2x-1=0

-4x=0

x=0

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Write the zero of the polynomial p(x) = 5x + 2.
Nookie1986 [14]

Answer:

-2/5

Step-by-step explanation:

p(x) = 5x + 2

plug p(x) = 0

5x + 2 = 0

5x = - 2

x = - 2/5

6 0
3 years ago
Read 2 more answers
Brooke has to set up 70 chairs in equal rows for the class talent show. But, there is not room for more then 20 rows. What are t
grin007 [14]

Answer:

The answer is 2, 5, 7, 10, and 14.


Step-by-step explanation:

Let x be the number of rows. There is not room for more than 20 rows:

x ≤ 20


Since all rows must be equal, when we divide 70 by x, we must get a whole number. So, we must find all divisors of 70 smaller or equal to 20 and they are: 2, 5, 7, 10, and 14.

So, the possibilities are:

2 rows with 35 chairs in each row

5 rows with 14 chairs in each row

7 rows with 10 chairs in each row

10 rows with 7 chairs in each row

14 rows with 5 chairs in each row

Hope this helps!!


~CoCo

7 0
3 years ago
HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! 14PTS!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
maksim [4K]

Answer:

a. d.

Step-by-step explanation:

4 0
3 years ago
Prove that the sum of the squares of any two odd numbers is always even
ValentinkaMS [17]

Answer:

proof below

Step-by-step explanation:

Remember that a number is even if it is expressed so n = 2k. It is odd if it is in the form 2k + 1 (k is just an integer)

Let's say we have to odd numbers, 2a + 1, and 2b + 1. We are after the sum of their squares, so we have (2a + 1)^2 + (2b + 1)^2. Now let's expand this;

(2a + 1)^2 + (2b + 1)^2 = 4a^2 + 4a + 4b + 4b^2 + 4b + 2

= 2(2a^2 + 2a + 2b^2 + 2b + 1)

Now the sum in the parenthesis, 2a^2 + 2a + 2b^2 + 2b + 1, is just another integer, which we can pose as k. Remember that 2 times any random integer, either odd or even, is always even. Therefore the sum of the squares of any two odd numbers is always even.

6 0
3 years ago
Problem<br> How many numbers between 111 and 100100100 (inclusive) are divisible by 101010 or 777?
olga2289 [7]

I suspect you meant

"How many numbers between 1 and 100 (inclusive) are divisible by 10 or 7?"

• Count the multiples of 10:

⌊100/10⌋ = ⌊10⌋ = 10

• Count the multiples of 7:

⌊100/7⌋ ≈ ⌊14.2857⌋ = 14

• Count the multiples of the LCM of 7 and 10. These numbers are coprime, so LCM(7, 10) = 7•10 = 70, and

⌊100/70⌋ ≈ ⌊1.42857⌋ = 1

(where ⌊<em>x</em>⌋ denotes the "floor" of <em>x</em>, meaning the largest integer that is smaller than <em>x</em>)

Then using the inclusion/exclusion principle, there are

10 + 14 - 1 = 23

numbers in the range 1-100 that are divisible by 10 or 7. In other words, add up the multiples of both 10 and 7, then subtract the common multiples, which are multiples of the LCM.

6 0
3 years ago
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