Respuesta :
we know that
If two numbers have a sum of zero, then we say they are additive inverses
so
case A)
[tex]x^{2} +3x-2[/tex]
[tex]-x^{2} -3x+2[/tex]
Sum the polynomials
[tex](x^{2} +3x-2)+(-x^{2} -3x+2)=0[/tex]
therefore
they are additive inverses
case B)
[tex]-y^{7} -10[/tex]
[tex]-y^{7} +10[/tex]
Sum the polynomials
[tex](-y^{7} -10)+(-y^{7} +10)=-2y^{7}[/tex]
[tex]-2y^{7}\neq 0[/tex]
therefore
they are not additive inverses
case C)
[tex]6z^{5} +6z^{5}-6z^{4}[/tex]
[tex](-6z^{5}) +(-6z^{5})+6z^{4}[/tex]
Sum the polynomials
[tex](6z^{5} +6z^{5}-6z^{4})+((-6z^{5}) +(-6z^{5})+6z^{4})=0[/tex]
therefore
they are additive inverses
case D)
[tex]x-1[/tex]
[tex]1-x[/tex]
Sum the polynomials
[tex](x-1)+(1-x)=0[/tex]
therefore
they are additive inverses
case E)
[tex](-5x^{2})+(-2x)+(-10)[/tex]
[tex]5x^{2}-2x+10[/tex]
Sum the polynomials
[tex]((-5x^{2})+(-2x)+(-10))+(5x^{2}-2x+10)=-4x[/tex]
[tex]-4x}\neq 0[/tex]
therefore
they are not additive inverses
Additive inverse means that the sum of two numbers will be zero and the options that are additive inverse is A), C), and D).
Additive inverse means that the sum of two numbers will be zero. Now, check all the given options:
A). [tex]x^2+3x-2\; ;\; -x^2-3x+2[/tex]
Sum of both the polynoimials will be:
[tex]=(x^2+3x-2)+(-x^2-3x+2)[/tex]
[tex]=0[/tex]
B). [tex]-y^7-10\; ;\; -y^7+10[/tex]
Sum of both the polynoimials will be:
[tex]=(-y^7-10)+(-y^7+10)[/tex]
[tex]=-2y^7[/tex]
C). [tex]6z^5+6z^5-6z^4\; ;\; -6z^5+(-6z^5)+6z^4[/tex]
Sum of both the polynoimials will be:
[tex]=6z^5+6z^5-6z^4+( -6z^5+(-6z^5)+6z^4)[/tex]
[tex]=0[/tex]
D). [tex]x-1\; ;\; 1-x[/tex]
Sum of both the polynoimials will be:
[tex]= x-1+1-x[/tex]
[tex]=0[/tex]
E). [tex](-5x^2)+(-2x)+(-10)\; ;\; 5x^2-2x+10[/tex]
Sum of both the polynoimials will be: [tex]=(-5x^2)+(-2x)+(-10)+ 5x^2-2x+10[/tex]
[tex]=-4x[/tex]
From the above calculation it can be concluded that option A), C), and D) are additive inverse.
For more information, refer the link given below:
https://brainly.com/question/19770987
