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Makovka662 [10]
3 years ago
14

Factor the polynomial completely. 6x2 - 3x + 2

Mathematics
2 answers:
Stella [2.4K]3 years ago
6 0

Answer:

option D

Step-by-step explanation:

Factor the polynomial completely. 6x^2 - 3x + 2

Apply AC method to factor

6 times 2 is 12. Product is 12 and sum is -3

We cannot find out two factors whose product is 12 and sum is -3.

Since we cannot find out two factors the given expression is prime.

It is not factorable . It cannot be factored.

Sliva [168]3 years ago
4 0
The answer is d because it cannot be factored
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If CB =3(x+8) and AD= 4(x+3), find AB
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AB=CD-CA-BD

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AB=CD-2(BC+CD)

AB=CD-2BC-2CD

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AB=-{98-6(x+8)}

AB= 6x-50

6 0
3 years ago
Which tables could be used to verify that the functions they represent are inverses of each other? Select two
devlian [24]
<h3>Answers:  C and D</h3>

Why? Because they are mirrors of each other. In other words, one table has its x and y row swapped compared to the other table.

The input x = -120 leads to the output y = 627 in table C. Then in table D, we have the input x = 627 to go the output y = -120. Refer to the first column of each table mentioned. The same applies for the other columns as well. The inverse table undoes everything the original table does.

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3 years ago
What is the value of 7x^2 + 3x + 4 when x=2?
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8 0
3 years ago
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For each of the following vector fields
olga nikolaevna [1]

(A)

\dfrac{\partial f}{\partial x}=-16x+2y

\implies f(x,y)=-8x^2+2xy+g(y)

\implies\dfrac{\partial f}{\partial y}=2x+\dfrac{\mathrm dg}{\mathrm dy}=2x+10y

\implies\dfrac{\mathrm dg}{\mathrm dy}=10y

\implies g(y)=5y^2+C

\implies f(x,y)=\boxed{-8x^2+2xy+5y^2+C}

(B)

\dfrac{\partial f}{\partial x}=-8y

\implies f(x,y)=-8xy+g(y)

\implies\dfrac{\partial f}{\partial y}=-8x+\dfrac{\mathrm dg}{\mathrm dy}=-7x

\implies \dfrac{\mathrm dg}{\mathrm dy}=x

But we assume g(y) is a function of y alone, so there is not potential function here.

(C)

\dfrac{\partial f}{\partial x}=-8\sin y

\implies f(x,y)=-8x\sin y+g(x,y)

\implies\dfrac{\partial f}{\partial y}=-8x\cos y+\dfrac{\mathrm dg}{\mathrm dy}=4y-8x\cos y

\implies\dfrac{\mathrm dg}{\mathrm dy}=4y

\implies g(y)=2y^2+C

\implies f(x,y)=\boxed{-8x\sin y+2y^2+C}

For (A) and (C), we have f(0,0)=0, which makes C=0 for both.

4 0
3 years ago
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