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3 years ago

5

If 2 marbles are drawn from the jar, one at a time, without replacing the first, what is the probability that the first one is r

ed and the second is yellow?

1 answer:

Hitman42 [59]3 years ago

6 0

Are there any numbers of how many there are of the different colors?

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Which expression is equivalent to (16 x Superscript 8 Baseline y Superscript negative 12 Baseline) Superscript one-half?.

loris [4]

To solve the problem we must know the Basic Rules of Exponentiation.

<h2>Basic Rules of Exponentiation</h2>

$x^ax^b = x^{(a+b)}$
$\dfrac{x^a}{x^b} = x^{(a-b)}$
$(a^a)^b =x^{(a\times b)}$
$(xy)^a = x^ay^a$

$x^{\frac{3}{4}} = \sqrt[4]{x^3}= (\sqrt[3]{x})^4$

The solution of the expression is $\dfrac{4x^4}{y^6}$ .

<h2>Explanation</h2>

Given to us

$(16x^8y^{12})^{\frac{1}{2}}$

Solution

We know that 16 can be reduced to $2^4$ ,

$=(2^4x^8y^{12})^{\frac{1}{2}}$

Using identity $(xy)^a = x^ay^a$ ,

$=(2^4)^{\frac{1}{2}}(x^8)^{\frac{1}{2}}(y^{12})^{\frac{1}{2}}$

Using identity $(a^a)^b =x^{(a\times b)}$ ,

$=(2^{4\times \frac{1}{2}})\ (x^{8\times\frac{1}{2}})\ (y^{12\times{\frac{1}{2}}})$

Solving further

$=2^2x^4y^{-6}$

Using identity $\dfrac{x^a}{x^b} = x^{(a-b)}$ ,

$=\dfrac{2^2x^4}{y^6}$

$=\dfrac{4x^4}{y^6}$

Hence, the solution of the expression is $\dfrac{4x^4}{y^6}$ .

Learn more about Exponentiation:

brainly.com/question/2193820

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2 years ago

IM GIVING 40 POINTS PLEASEEEEE

Mars2501 [29]

Answer:

Answer is $110.25. You figure out how.

Step-by-step explanation:

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2 years ago

What is the solution for <br> 10+2w>=22 or 5w-8>-12

Marina CMI [18]

Answer:

no solution

Step-by-step explanation:

10+2w>=22 or 5w-8>-12

10-2w>=22 or (5w-2w)-8>-12

10>=22 or 3w+8>12

18>=22 or 3w>12

18/3>=22 or 3w/3>12

6>=22 or w>12

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2 years ago

Please help me the photo is my queston

Studentka2010 [4]

Answer:

i think it would be 30

Step-by-step explanation:

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3 years ago

Where is the hole for the following function located f(x)= x+3/(x-4)(x+3)

Marizza181 [45]

In a rational function, holes are located where f(x) is undefined(the denominator is 0), but the undefined points can be factored out. In f(x), when x is 4 or -3, f(x) is undefined. But, x+3 can be factored out, so it is a hole. The hole is located at (-3,-1/7)

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3 years ago

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