What is the source of carbon in fossil fuels? *

1 answer:

6 0

Decomposing plants and other organisms, buried beneath layers of sediment and rock, have taken millennia to become the carbon-rich deposits we now call fossil fuels

You might be interested in

Hey guy pls help <br> explain answer <br> no links

laiz [17]

Answer:

4, 5

Step-by-step explanation:

the mode is 4 since mode is the number/s that appear most. Median is the number in the middle when the range is in order. That is 5

8 0

3 years ago

Pleaseee someone help me in this question ihave bben eating for half an hour and no one helped me.

pav-90 [236]

Cross Multiply by 'x - 5'
Hence
y(x - 5) = 4 -3x
xy - 5y = 4 - 3x

Move all the terms containing an 'x' to one side of the equals and all the others to the opposite side.
xy + 3x = 4 + 5y
Factor out 'x'
x(y + 3) = 4 + 5y
'Cross' divide by 'y + 3'
x = ( 4 + 5y) / (y + 3 )

'x' is now the subject.

8 0

3 years ago

Factor completely x^8- 16.

Nataly_w [17]

Answer:

Step-by-step explanation:

So in this example we'll be using the difference of squares which essentially states that: $(a-b)(a+b)=a^2-b^2$ or another way to think of it would be: $a-b=(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})$ . So in this example you'll notice both terms are perfect squares. in fact x^n is a perfect square as long as n is even. This is because if it's even it can be split into two groups evenly for example, in this case we have x^8. so the square root is x^4 because you can split this up into (x * x * x * x) * (x * x * x * x) = x^8. Two groups with equal value multiplying to get x^8, that's what the square root is. So using these we can rewrite the equation as:

$x^8-16 = (x^4-4)(x^4+4)$

Now in this case you'll notice the degree is still even (it's 4) and the 4 is also a perfect square, and it's a difference of squares in one of the factors, so it can further be rewritten:

$x^4-4 = (x^2-2)(x^2+2)$

So completely factored form is: $(x^2-2)(x^2+4)(x^4+4)$

I'm assuming that's considered completely factored but you can technically factor it further. While the identity difference of squares technically only applies to difference of squares, it can also be used on the sum of squares, but you need to use imaginary numbers. Because $x^2+4 = x^2-(-4)$ . and in this case a=x^2 and b=-4. So rewriting it as the difference of squares becomes: $x^4+4 = x^4 - (-4) = (x^2-\sqrt{-4})(x^2+\sqrt{-4}) = (x^2-2i)(x^2+2i)$ just something that might be useful in some cases.