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3 years ago

A magazine has 96 pages. there are advertisements on 75% of those pages. how many pages do not have advertisements?

Mathematics

1 answer:

Diano4ka-milaya [45]3 years ago

8 0

If you find out the other 25% then that should give you your answer ;) hope this helps you

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Which word should replace the question mark (?) in the

artcher [175]

Answer:

speed I'm pretty sure sorry if its wrong

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3 years ago

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Rick recorded the information shown below. What type of graph would be the most appropriate for Rick to use

Marina86 [1]

A Line graph

Sure hope this helps you

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2 years ago

I know this is a supervised data mining task because (choose all that apply) A. Because the target belonged to the same group as

Vinvika [58]

Answer:

The answer to your question is C. There a specific quantifiable target that we are interested in or trying to predict.

Step-by-step explanation:

I hope this helps and have a good day!

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2 years ago

Calculate the average rate of change for the given function, from x = −3 to x = 7.

emmainna [20.7K]

Answer:

The average rate of change for the given function, from x = −3 to x = 7 will be:

Average rate = -5

Step-by-step explanation:

Given the table

x f(x)

-3 50

3 10

7 0

As the interval is from x = -3 to x = 7.

at x₁ = -3, f(x₁) = 50

at x₂ = 7, f(x₂) = 0

Using the formula to determine the average rate of change at which the total cost increases will be:

Average rate = [f(x₂) - f(x₁)] / [ x₂ - x₁]

= [0 - 50] / [7-(-3)]

= [-50] / [7+3]

= -50 / 10

= -5

Therefore, the average rate of change for the given function, from x = −3 to x = 7 will be:

Average rate = -5

6 0

3 years ago

Which expression is a sum of cubes?

erik [133]

we know that

A polynomial in the form $a^{3} +b^{3}$ is called a sum of cubes

Let's verify each case to determine the solution

case A) $-64x^{6} y^{12} +125x^{16} y^{3}$

we know that

$-64=-4^{3}$

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

$y^{12}= (y^{4})^{3}$

$125=5^{3}$

$x^{16}=x^{15} *x=x*(x^{5})^{3}$ -------> is not a perfect cube

$y^{3}= (y)^{3}$

therefore

the case A) is not a sum of cubes

case B) $-32x^{6} y^{12} +125x^{16} y^{3}$

we know that

$-32=-2^{5}$ -------> is not a perfect cube

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

$y^{12}= (y^{4})^{3}$

$125=5^{3}$

$x^{16}=x^{15} *x=x*(x^{5})^{3}$ -------> is not a perfect cube

$y^{3}= (y)^{3}$

therefore

the case B) is not a sum of cubes

case C) $32x^{6} y^{12} +125x^{9} y^{3}$

we know that

$32=2^{5}$ -------> is not a perfect cube

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

$y^{12}= (y^{4})^{3}$

$125=5^{3}$

$x^{9}=(x^{3})^{3}$

$y^{3}= (y)^{3}$

therefore

the case C) is not a sum of cubes

case A) $64x^{6} y^{12} +125x^{9} y^{3}$

we know that

$64=4^{3}$

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

$y^{12}= (y^{4})^{3}$

$125=5^{3}$

$x^{9}=(x^{3})^{3}$

$y^{3}= (y)^{3}$

Substitute

$4^{3}(x^{2})^{3}(y^{4})^{3} +5^{3}(x^{3})^{3}(y)^{3}$

$(4x^{2}y^{4})^{3} +(5x^{3}y)^{3}$

therefore

the answer is

$64x^{6} y^{12} +125x^{9} y^{3}$ is a sum of cubes

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3 years ago

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