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3 years ago

How much of the sphere do you have to paint red so that no matter ow you put a cube in it, some corner touches red?

Mathematics

1 answer:

Triss [41]3 years ago

4 0

Explanation:

To achive that you have to piant a ring in red, which will have its big diameter equal to the sphere diameter and its small diameter equal to the cube diagonal.

The diagonal of the cube can be calculated using Pithagoras:

$D^2=L^2+L^2$

Where D is the diagonal and L is the side of the cube

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10 minutes to answer Write an expression equivalent to 4 - (-7) help

Mama L [17]

11 because -7 subtracted by 4 add kore to the 7 so think of it this way -7 is a bad guy and 4 is a bad guy to because its a subtraction so if both bad guys combine you get 11 because a negative with negative makes a positive so its 11

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3 years ago

Anyone want to ch at?

algol13

Answer:

sure?

Step-by-step explanation:

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3 years ago

Read 2 more answers

Find the critical numbers of the function. (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE

EleoNora [17]

Answer:

0, 10

Step-by-step explanation:

The given function is:

$g(y) = \frac{y-5}{y^2-3y+15}$

According to the quotient rule:

$d(\frac{f(y)}{h(y)}) = \frac{f(y)*h'(y)-h(y)*f'(y)}{h^2(y)}$

Applying the quotient rule:

$g(y) = \frac{y-5}{y^2-3y+15}\\g'(y)=\frac{(y-5)*(2y-3)-(y^2-3y+15)*(1)}{(y^2-3y+15)^2}$

The values for which g'(y) are zero are the critical points:

$g'(y)=0=\frac{(y-5)*(2y-3)-(y^2-3y+15)*(1)}{(y^2-3y+15)^2}\\(y-5)*(2y-3)-(y^2-3y+15)=0\\2y^2-3y-10y+15-y^2+3y-15\\y^2-10y=0\\y=\frac{10\pm \sqrt 100}{2}\\y_1=\frac{10-10}{2}= 0\\y_2=\frac{10+10}{2}=10$

The critical values are y = 0 and y = 10.

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3 years ago

Bob is on the school swim team and has practice 4 days per week. He has to practice twice per day, in the morning and in the eve

mylen [45]

Answer:

1h every morning

Step-by-step explanation:

3 0

3 years ago

What’s the answer???

stellarik [79]

Answer: (3,4)

Each input can only have one output for a function

The other pairs give a different output for the input so that would make it a relation not a function.

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4 years ago