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3 years ago

Identifying Additive Inverses

Mathematics

1 answer:

mestny [16]3 years ago

4 0

Answer:

he additive inverse of:

a) $-6x^2-x-2$ is : $6x^2+x+2$

b) $6x^2-x+2$ is : $-6x^2+x-2$

c) $6x^2+x-2$ is : $-6x^2-x+2$

d) $6x^2+x+2$ is : $-6x^2-x-2$

Step-by-step explanation:

You need to consider that the additive inverse of a polynomial is that polynomial that consists of the opposite of each term of the polynomial given.

Then, the additive inverse of:

a) $-6x^2-x-2$ is : $6x^2+x+2$

b) $6x^2-x+2$ is : $-6x^2+x-2$

c) $6x^2+x-2$ is : $-6x^2-x+2$

d) $6x^2+x+2$ is : $-6x^2-x-2$

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Y_Kistochka [10]

Answer:

$y=\frac{5}{4}x-3$

Step-by-step explanation:

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3 years ago

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irga5000 [103]

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3 years ago

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pictured below are stacks of solid cubes. determine the number of cubes in the stack and the number of faces glued together.

Iteru [2.4K]

Solution

For this case we need to find the number of cubes in the stack given so then we can do this:

The first line present 5 cubes

The second line present 5 cubes

And the last line we got:

10 cubes

Then the total of cubes are:

10 +5+5 = 20 cubes

If we need to find the number of faces we can do the following:

5+ 4+ 4+4 +4 +1 + (5+4+4+4+4+1 ) + (5+4+4+4+4+1 ) + (5+ 4+4+4 +4 )

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1 year ago

Solve the following systems by Elimination

Alenkasestr [34]

Answer:

The solutions for both system of equations are as follows:

(5,2)
(2,-1)

Step-by-step explanation:

The first set of equations is:

$4x+6y=32\\3x-6y=3\\$

It can clearly be seen that the coefficients of y are already same in magnitude with different signs so we have to add both equations

So adding both equations, we get

$4x+6y+3x-6y = 32+3\\7x = 35\\\frac{7x}{7} = \frac{35}{7}\\x = 5$

Putting x=5 in equation 1

$4(5)+6y = 32\\20+6y = 32\\6y = 32-20\\6y = 12\\\frac{6y}{6} = \frac{12}{6}\\y = 2$

The solution is (5,2)

The second set of simultaneous equations is:

$-3x+5y=-113x+7y=-1$

We can see that the coefficients of x in both equations are same in magnitude with opposite signs so

Adding both equations

$-3x+5y+3x+7y = -11-1\\12y = -12\\\frac{12y}{12} = \frac{-12}{12}\\y = -1$

Putting y= -1 in first equation

$-3x+5(-1)=-11\\-3x-5=-11\\-3x=-11+5\\-3x=-6\\\frac{-3x}{-3} = \frac{-6}{-3}\\x = 2$

The solution is: (2,-1)

Hence,

The solutions for both system of equations are as follows:

(5,2)
(2,-1)

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3 years ago

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Julli [10]

I think the answer is.e=mc3

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3 years ago