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Nonamiya [84]
2 years ago
7

The table shown below lists the time, in days, that it takes for

Mathematics
1 answer:
Crazy boy [7]2 years ago
5 0

Answer:gotchu g there it is

Step-by-step explanation:

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On the number line below, the letters a and b are the same distance from 0. What is a + b?
jolli1 [7]

Answer:

The answer here would be 0.

Step-by-step explanation:

If a and b are both equal distances from 0, then adding the together would equal 0. Since a is the inverse of b,  using any number and its inverse will show you why the answer is 0. For example, if b is 9 then a would be -9. Adding these together equals 0.

5 0
3 years ago
Solve 3x^2 6x=12 by quadratic formula
kari74 [83]
That would be x=0.87358
4 0
3 years ago
1/4 to 1/2 find the percent of change-increase or decrease
Vika [28.1K]
The percent of change would be -33.3% which would then equal a 33.3% decrease (when decimal is rounded) 

what I did was turn 1/4 and 1/2 into decimals to make it easier to solve 

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3 years ago
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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

8 0
2 years ago
I need help please…thank youuu!!!
DochEvi [55]

Answer:

I will help ukshagaakavlabsbakakahsjskmdlslsbsbs skshsvs

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