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Vadim26 [7]
3 years ago
9

Ruth has 6 chores to do. If each chore does not depend on the others, in how many different orders could she do the chores?

Mathematics
2 answers:
Dafna11 [192]3 years ago
7 0
The answer is above you shoukd mark it brainliest
Mice21 [21]3 years ago
3 0
First chore, 6 to choose from.
Second chore, 5 to choose from.
Third chore, 4 to choose from.
Fourth chore, 3 to choose from.
Fifth chore, 2 to choose from.
Sixth chore, 1 to "choose" from.
The total number of orders possible is the product
6*5*4*3*2*1=6!=720
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PLEASEE HELP!! what is the perimeter of x?? HURRYYYY
Leno4ka [110]
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3 years ago
HELP!!!!!! I believe its B but I am not sure! :(
n200080 [17]

Answer:

A

Step-by-step explanation:

When solving for x as an exponent, we need to use logarithms in order to undo the operation and rearrange the terms. We use log rules to bring down the exponent and solve. Logarithms are the inverse operations to exponents and vice versa. We have one special kind of logarithm called the natural logarithm whose base is e. We write it as ln. Since our base is e here, we will use the natural logarithm to rearrange and isolate x.

e^{4x-1} =3

We begin by applying the natural logarithm to each side.

ln(e^{4x-1}) =ln(3)

Log rules allow use to rearrange the exponent as multiplication in front of the log.

(4x-1)ln(e) =ln(3)

ln e as an inverse simplifies to 1.

(4x-1)(1)=ln(3)

We now apply the inverse operations for subtraction and multiplication.

4x-1+1=1+ln(3)\\4x=1+ln(3)\\\frac{4x}{4} =\frac{1+ln3}{4} \\x =\frac{1+ln3}{4}

Option A is correct.

5 0
3 years ago
Find a degree 3 polynomials with real coefficient having zeros 2 and 4i and a leading coefficient of 1. Write P in expanded form
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7 0
3 years ago
Can someone help me find the equivalent expressions to the picture below? I’m having trouble
miss Akunina [59]

Answer:

Options (1), (2), (3) and (7)

Step-by-step explanation:

Given expression is \frac{\sqrt[3]{8^{\frac{1}{3}}\times 3} }{3\times2^{\frac{1}{9}}}.

Now we will solve this expression with the help of law of exponents.

\frac{\sqrt[3]{8^{\frac{1}{3}}\times 3} }{3\times2^{\frac{1}{9}}}=\frac{\sqrt[3]{(2^3)^{\frac{1}{3}}\times 3} }{3\times2^{\frac{1}{9}}}

           =\frac{\sqrt[3]{2\times 3} }{3\times2^{\frac{1}{9}}}

           =\frac{2^{\frac{1}{3}}\times 3^{\frac{1}{3}}}{3\times 2^{\frac{1}{9}}}

           =2^{\frac{1}{3}}\times 3^{\frac{1}{3}}\times 2^{-\frac{1}{9}}\times 3^{-1}

           =2^{\frac{1}{3}-\frac{1}{9}}\times 3^{\frac{1}{3}-1}

           =2^{\frac{3-1}{9}}\times 3^{\frac{1-3}{3}}

           =2^{\frac{2}{9}}\times 3^{-\frac{2}{3} } [Option 2]

2^{\frac{2}{9}}\times 3^{-\frac{2}{3} }=(\sqrt[9]{2})^2\times (\sqrt[3]{\frac{1}{3} } )^2 [Option 1]

2^{\frac{2}{9}}\times 3^{-\frac{2}{3} }=(\sqrt[9]{2})^2\times (\sqrt[3]{\frac{1}{3} } )^2

                =(2^2)^{\frac{1}{9}}\times (3^2)^{-\frac{1}{3} }

                =\sqrt[9]{4}\times \sqrt[3]{\frac{1}{9} } [Option 3]

2^{\frac{2}{9}}\times 3^{-\frac{2}{3} }=(2^2)^{\frac{1}{9}}\times (3^{-2})^{\frac{1}{3} }

               =\sqrt[9]{2^2}\times \sqrt[3]{3^{-2}} [Option 7]

Therefore, Options (1), (2), (3) and (7) are the correct options.

6 0
2 years ago
What scale factor was used when going from the grey triangle to the yellow triangle
sertanlavr [38]

Answer:

Scale factor of 1/3

Step-by-step explanation:

3 0
3 years ago
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