2 years ago

Cheek your level in mathematics

1 answer:

7 0

Sorry, I can't help you as I don't know this

This concept has not came yet in my class that's why I don't know

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baherus [9]

Hello 16,700 is left over

8 0

3 years ago

Zina [86]

(A) Product of $-2x^{3}+x-5$ and $x^{3}-3x-4$ is:

$(-2x^{3}+x-5)(x^{3}-3x-4)$

$=-2x^{3}(x^{3}-3x-4)+x(x^{3}-3x-4)-5(x^{3}-3x-4)$

$=(-2x^{3})(x^{3})+(-2x^{3})(-3x)+(-2x^{3})(-4)+x(x^{3})+(x)(-3x)+(x)(-4)+(-5)(x^{3})+(-5)(-3x)+(-5)(-4)$

$=-2x^{6}+6x^{4}+8x^{3}+x^{4}-3x^{2}-4x-5x^{3}+15x+20$

$=-2x^{6}+6x^{4}+x^{4}+ 8x^{3}-5x^{3}-3x^{2}-4x+15x+20$

$=-2x^{6}+7x^{4}+3x^{3}-3x^{2}+11x+20$

(B) Yes.

Product of $-2x^{3}+x-5$ and $x^{3}-3x-4$ = Product of $x^{3}-3x-4$ and $-2x^{3}+x-5$ because multiplication is commutative.

Commutative Property of multiplication says that a.b = b.a.

Thus, multiplication is same irrespective of the order of two numbers.

5 0

3 years ago

GenaCL600 [577]

20t^2+9t-20 is the answer

7 0

2 years ago

Vesna [10]

B

If he can type 1/3 as many words as her that would be 1/3j which is the same as j/3.

7 0

3 years ago

kozerog [31]

1. -3x -20
2. -6x-16
3. -14-x

6 0

3 years ago