3 years ago

Can someone please help me as soon as possible thanks

1 answer:

exis [7]3 years ago

7 0

Left row 1. Supplementary 2.complementary 3 supplementary
Right row 1. Complementary 2 supplementary 3 supplementary

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alukav5142 [94]

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Egg

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morpeh [17]

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Marat540 [252]

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The answer is D. (8•4 ) + (2•4) + 2

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3 years ago

andreev551 [17]

Notice that you have a difference of cubes:

$\dfrac{x^3}8-\dfrac{y^3}{27}=\left(\dfrac x2\right)^3-\left(\dfrac y3\right)^3$

which can be factored using the rule,

$a^3-b^3=(a-b)(a^2+ab+b^2)$

So,

$\dfrac{x^3}8-\dfrac{y^3}{27}=\left(\dfrac x2-\dfrac y3\right)\left(\dfrac{x^2}4+\dfrac{xy}6+\dfrac{y^2}9\right)$

and just to make things look nicer, we can pull out 1/6 from the first factor and 1/36 from the second to get

$\dfrac{x^3}8-\dfrac{y^3}{27}=\dfrac1{216}(3x-2y)(9x^2+6xy+4y^2)$

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4 years ago

tatyana61 [14]

Yes it is. log(1) and ln(1) both equal zero. But you cannot take the logarithm of 0.

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4 years ago