You have a bag of marbles that has 3 blue marbles and 4 red ones. What is the chance that you pick a blue OR a red marble?

Mathematics

1 answer:

nikdorinn [45]2 years ago

5 0

Answer:

7/7

Step-by-step explanation:

Add up the amount of the marbles in total and read what color the marble they ask for the chances of. The color of marble should go first then the total of the marbles. (they could ask for more than one color so just add up both the colors given.)

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Which shows the expression x^2-1/x^2-x in simplest form

____ [38]

Answer:

$\large\boxed{\dfrac{x+1}{x}=1+\dfrac{1}{x}}$

Step-by-step explanation:

$\dfrac{x^2-1}{x^2-x}=(*)\\\\x^2-1=x^2-1^2\qquad\text{use}\ a^2-b^2=(a-b)(a+b)\\=(x-1)(x+1)\\\\x^2-x=(x)(x)-(x)(1)\qquad\text{use the distributive property}\\=x(x-1)\\\\(*)=\dfrac{(x-1)(x+1)}{x(x-1)}\qquad\text{cancel}\ (x-1)\\\\=\dfrac{x+1}{x}=\dfrac{x}{x}+\dfrac{1}{x}=1+\dfrac{1}{x}$

4 0

3 years ago

Read 2 more answers

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.