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3 years ago

What is the sign of 3x y3xy3, x, y when x>0x>0x, is greater than, 0 and y<0y<0y, is less than, 0?

Mathematics

2 answers:

jekas [21]3 years ago

4 0

Answer:

$Negative$

Step-by-step explanation:

$We\ have\ to\ find\ sign\ of\ 3xy^3\\\\x>0\ and\ y0,y$

Vladimir [108]3 years ago

3 0

Answer:

Negative

Step-by-step explanation:

Positive x positive= positive

Positive x negative= negative

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Expand the given power by using Pascal’s triangle. (9a - 10b)^6

xxMikexx [17]

Answer:

$531441a^6-3542940a^5b+9841500a^4b^2-14580000a^3b^3+12150000a^2b^4-5400000ab^5+1000000b^6$

Step-by-step explanation:

1 n=0

1 1 n=1

1 2 1 n=2

1 3 3 1 n=3

1 4 6 4 1 n=4

1 5 10 10 5 1 n=5

1 6 15 20 15 6 1 n=6

This is where n is the exponent in

$(x+y)^n$ .

$(x+y)^6=1x^6+6x^5y+15x^4y^2+20x^3y^3+15x^2y^4+6xy^5+1y^6$

Now we want to expand:

$(9a-10b)^6$ or we we can rewrite as $(9a+(-10b))^6$ .

Let's replace $x$ with $(9a)$ and $y$ with $(-10b)$ in the expansion:

$(x+y)^6=1x^6+6x^5y+15x^4y^2+20x^3y^3+15x^2y^4+6xy^5+1y^6$

$((9a)+(-10b))^6$

$=1(9a)^6+6(9a)^5(-10b)+15(9a)^4(-10b)^2+20(9a)^3(-10b)^3+15(9a)^2(-10b)^4+6(9a)(-10b)^5+1(-10b)^6$

Let's simplify a bit:

$=9^6a^6-60(9)^5a^5b+15(-10)^2(9)^4a^4b^2+20(9)^3(-10)^3a^3b^3+15(9)^2(-10)^4a^2b^4+6(9)(-10)^5ab^5+(-10)^6b^6$

$=531441a^6-3542940a^5b+9841500a^4b^2-14580000a^3b^3+12150000a^2b^4-5400000ab^5+1000000b^6$

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