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zaharov [31]
1 year ago
15

A number cube labeled 1 through 6 is rolled.

Mathematics
1 answer:
lbvjy [14]1 year ago
8 0

Answer: 3/6 or 1/2 or 50%

Step-by-step explanation: 1, 2, 3 4, 5, 6 are each face

2, 4, and 6 are even.

3 out of the 6 are even (3/6)

reduce (1/2)

1/2 = 50%

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PLEASE HELP ASAP!!! I don’t know how to do this problem or where to start! How do I solve this?
Leto [7]

Answer:

W = 4.95

Step-by-step explanation:

You want to start by writing down what you know, and forming a system of equations.

L= length W= width

2L+2W=14.7    

L= 2.4

On the left side of the equation, you're adding all your side lengths, and on the right, is the total perimeter. (Also could be written L+L+W+W = 14.7)

You would then substitute L from the bottom equation into the top equation to get:

2(2.4) +2W=14.7

Solving:

4.8+2w=14.7

W= 4.95

To check your answer simply add all the sides together and make sure it equals your perimeter. You can also plug W and L back into the original equation.

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3 years ago
Which number line represents the solution set for inequality 3x<-9 ?
vovikov84 [41]

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Read 2 more answers
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marshall27 [118]

Answer:

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Step-by-step explanation:

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8 0
3 years ago
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What’s is the answer
kvasek [131]

When you are multiplying an exponent directly into a number/variable with an exponent, you multiply the exponents together.

For example:

(x^{2} )^{3} = x^6

(x^{3} )^5=x^{15}


When you are multiplying a variable with an exponent by another variable with an exponent, you add the exponents together.

For example:

(x^{2} )(x^{3})=x^{5}

(x^{1} )(x^{2})=x^{3}


(\frac{(x^{-3})(y^{2})}{(x^{4})(y^{6})} )^{3}=\frac{(x^{-9})(y^{6})}{(x^{12})(y^{18})}

You multiply 3 into each exponent in the numerator and the denominator

\frac{(x^{-9})(y^{6})}{(x^{12})(y^{18})}= \frac{y^{6}}{(x^{9})(x^{12})(y^{18})}

When you have a negative exponent, you move it to the other side of the fraction to make the exponent positive.

\frac{y^{6}}{(x^{21})(y^{18})} = \frac{1}{(x^{21})(y^{12})}


When you have something like this:

\frac{x^{2}}{x^5}

You subtract the exponents together, so:

\frac{x^2}{x^5} = x^{2-5} = x^{-3} = \frac{1}{x^3}


Your answer is the second option

3 0
3 years ago
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Vesna [10]

Answer:

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6 0
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