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allsm [11]
3 years ago
14

What is the quadratic term for f(x) = 2x(x - 1) - 2x?

Mathematics
1 answer:
erica [24]3 years ago
6 0
I believe it would be 6. the first number and exponent are usually your quadratic terms
You might be interested in
4m-2+(-8m)<br> Need answer ASAP please help me
DENIUS [597]

Answer:

-4m-2 Hope I helped

Step-by-step explanation: Remove parentheses.

4m−2−8m

Subtract 8m

from 4m

.

−4m−2

4 0
2 years ago
Read 2 more answers
What is 7999 +9100=?????????
Mama L [17]

the answer is 17099

6 0
3 years ago
The school store opened on the first day of school with 45 notebooks and 15 pencils. Within two days it sold all of these items.
malfutka [58]

The number of pencils sold on the first day  is 5 pencils.

The number of notebooks sold on the first day  is 10 notebooks.

The number of pencils sold on the second day 10 pencils.

The number of notebooks sold on the second day   35 notebooks.

<u>Step-by-step explanation:</u>

Given data,

  • The total number of notebooks in the store = 45 notebooks.
  • The total number of Pencils in the store = 15 pencils.
  • The number of days that all items were sold = 2 days.

Now, you have to calculate the no.of notebooks and no. of pencils sold each day.

<u>In the first day :</u>

The number of sale of notebooks and pencils are given by,

Twice as many notebooks were sold as pencils.

Let us take, the number of pencils sold on the first day  = x

And,  the number of notebooks sold on the first day  = 2x (Twice as pencils).

<u>In the second day :</u>

The number of sale of notebooks and pencils are given by,

For every 7 notebooks​ sold, 2 pencils were sold.

The number of pencils sold on the second day =   2y

The number of notebooks sold on the second day   = 7y

<u> The equation is framed for number of notebooks sold on each day :</u>

The number of notebooks sold ⇒ 2x + 7y = 45   -------(1)

<u> The equation  is framed for number of pencils sold on each day :</u>

The number of pencils sold: x + 2y = 15  ----------(2)

Solving the equations by multiplying eq(2) by 2 and subtract it from eq(1),

  2x + 7y = 45

-<u> (2x + 4y) = 30</u>

   <u>        3y = 15  </u>

⇒ y = 15/3

⇒ y = 5

The value of y is 5.

Substitute y=5 in eq (2),

⇒ x + 2(5) = 15

⇒ x + 10 = 15

⇒ x = 15 - 10

⇒ x = 5

The value of x is 5.

First day sale,

The number of pencils sold on the first day  = x ⇒ 5 pencils

The number of notebooks sold on the first day  = 2x ⇒ 10 notebooks

Second day sale,

The number of pencils sold on the second day =   2y ⇒ 10 pencils

The number of notebooks sold on the second day   = 7y ⇒ 35 notebooks.

8 0
2 years ago
What is the expression in factored form? 4x^2+11x+6<br><br>Show work please!!
Irina18 [472]
To solve this I'm going to split the middle term.
First multiply the first and last terms:
24x^2
So find two numbers that multiply to 24x^2 and add to 11x.
This would be 3x and 8x
Rewrite the problem as
4x^2+3x+8x+6
Take the first and 3rd and 2nd and 4th terms
4x^2 and 8x
and
3x and 6
Factor by grouping
Take out a 4x for the first group to get 4x(x+2)
Take out a 3 for the 2nd group to get 3(x+2)
Rewrite as (4x+3)(x+2)
Hope this helps.

6 0
3 years ago
Read 2 more answers
Use the Polynomial Identity below to help you create a list of 10 Pythagorean Triples:
Stolb23 [73]

Answer:

(3,4,5)

(6,8,10)

(5,12,13)

(8,15,17)

(12,16,20)

(7,24,25)

(10,24,26)

(20,21,29)

(16,30,34)

(9,40,41)

Just choose 2 numbers from {1,2,3,4,5,6,7,8,...} and make sure the one you input for x is larger.

Post the three in the comments and I will check them for you.

Step-by-step explanation:

We need to choose 2 positive integers for x and y where x>y.

Positive integers are {1,2,3,4,5,6,7,.....}.

I'm going to start with (x,y)=(2,1).

x=2 and y=1.

(2^2+1^2)^2=(2^2-1^2)^2+(2\cdot2\cdot1)^2

(4+1)^2=(4-1)^2+(4)^2

(5)^2=(3)^2+(4)^2

So one Pythagorean Triple is (3,4,5).

I'm going to choose (x,y)=(3,1).

x=3 and y=1.

(3^2+1^2)^2=(3^2-1^2)^2+(2\cdot3\cdot1)^2

(9+1)^2=(9-1)^2+(6)^2

(10)^2=(8)^2+(6)^2

So another Pythagorean Triple is (6,8,10).

I'm going to choose (x,y)=(3,2).

x=3 and y=2.

(3^2+2^2)^2=(3^2-2^2)^2+(2\cdot3\cdot2)^2

(9+4)^2=(9-4)^2+(12)^2

(13)^2=(5)^2+(12)^2

So another is (5,12,13).

I'm going to choose (x,y)=(4,1).

(4^2+1^2)^2=(4^2-1^2)^2+(2\cdot4\cdot1)^2

(16+1)^2=(16-1)^2+(8)^2

(17)^2=(15)^2+(8)^2

Another is (8,15,17).

I'm going to choose (x,y)=(4,2).

(4^2+2^2)^2=(4^2-2^2)^2+(2\cdot4\cdot2)^2

(16+4)^2=(16-4)^2+(16)^2

(20)^2=(12)^2+(16)^2

We have another which is (12,16,20).

I'm going to choose (x,y)=(4,3).

(4^2+3^2)^2=(4^2-3^2)^2+(2\cdot4\cdot3)^2

(16+9)^2=(16-9)^2+(24)^2

(25)^2=(7)^2+(24)^2

We have another is (7,24,25).

You are just choosing numbers from the positive integer set {1,2,3,4,... } and making sure the number you plug in for x is higher than the number for y.

I will do one more.

Let's choose (x,y)=(5,1).

(5^2+1^2)^2=(5^2-1^2)^2+(2\cdot5\cdot1)^2

(25+1)^2=(25-1)^2+(10)^2

(26)^2=(24)^2+(10)^2

So (10,24,26) is another.

Let (x,y)=(5,2).

(5^2+2^2)^2=(5^2-2^2)^2+(2\cdot5\cdot2)^2

(25+4)^2=(25-4)^2+(20)^2

(29)^2=(21)^2+(20)^2

So another Pythagorean Triple is (20,21,29).

Choose (x,y)=(5,3).

(5^2+3^2)^2=(5^2-3^2)^2+(2\cdot5\cdot3)^2

(25+9)^2=(25-9)^2+(30)^2

(34)^2=(16)^2+(30)^2

Another Pythagorean Triple is (16,30,34).

Let (x,y)=(5,4)

(5^2+4^2)^2=(5^2-4^2)^2+(2\cdot5\cdot4)^2

(25+16)^2=(25-16)^2+(40)^2

(41)^2=(9)^2+(40)^2

Another is (9,40,41).

5 0
3 years ago
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