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lutik1710 [3]
4 years ago
9

You’re baking a circular cake for a party. Your cake can be any size you want. What will the radius of your cake be and how many

slices will you be able to cut?

Mathematics
1 answer:
REY [17]4 years ago
7 0
What will the radius of your cake be?

This is a problem of geometry. Given that the cake is circular, the greater the cake the greater the radius of it. So, as shown in the figure 1, the radius will be the distance from the center to any extreme point of the circle.

How many slices will you be able to cut?

The total area of a circle is given by:

A= \pi  r^{2}

We need to fin how many slices will be cut, so let's calculate the area of the circular sector which can be obtained simply applying rule of three, so:

A_{cs}= \pi r^{2}( \frac{\theta}{2\pi})= \frac{r^{2}\theta}{2}

Let's name n the number of slices, if we divide the total area by n this result, each area  must equal, then:

\frac{\pi r^{2}}{n}=\frac{r^{2}\theta}{n}

Finally, we will be able to cut:

n= \frac{2\pi}{\theta} Slices

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Is the table shown proportional?Explain
ioda

Answer:

the first one is proportional because you could multiply by 4

6 0
3 years ago
A function is given. (a) Find all the local maximum and minimum values of the function and the value of x at which each occurs.
Allisa [31]

Answer:

a)

x = -1/√3 = -0.58    is a local maximum

x = 1/√3  =  0.58     is a local minimum

b)

(-1/√3 , 1/√3 ) DECREASING

(-∞, -1/√3) U (1/√3, ∞)    INCREASING

Step-by-step explanation:

To answer all of the questions we must obtain the derivative of the fucntion:

If

U(x) = 4(x^3 - x)

then

U'(x) = 4(3x^2 - 1)

U''(x) = 4(3*2 x) = 24 x

U'''(x) = 24

The local maxima and minima of the function U(x) can be found when U'(x) = 0

this occurs when :

3x^2 = 1

that is:

x = ±1/√3

We will know if they are a minimum or a maximum evaluating this points on the second derivative (you can look for it as <u><em>Second derivative test</em></u>), if the result is positive the point corresponds to a minimum and if it is negative it will be a maximum.

Here it is easy to determine wheather is a maximum or a minimum because the second derivative is 24x, therefore if x is negative or positive so the second derivative will be.

a)

x = -1/√3 = -0.58    is a local maximum

x = 1/√3  =  0.58     is a local minimum

b)

the intervals at which the function is increasing { decreasing } is given when the first derivative is positive { negative }

The first derivative will be positive when:

3x^2 > 1

|x| > 1/√3  -->  x > 1/√3   and    x < -1/√3

The first derivatice will be negative when:

3x^2 < 1

|x| < 1/√3  -->  x < 1/√3   and    x > -1/√3

Therefore the intervals are:

(-1/√3 , 1/√3 ) DECREASING

(-∞, -1/√3) U (1/√3, ∞)    INCREASING

<u><em>** The attached image is a plot of the fucntion where you can see part of the intervals and the local maximum and minimum</em></u>

8 0
3 years ago
I need help what is y over 5&gt;= 8
nadezda [96]

Answer:

3 i think

Step-by-step explanation:

8 0
3 years ago
If f (x) equals 3 + x / x - 3 what is f (a + 2)?​
xenn [34]

ANSWER

f(a + 2) =  \frac{a + 5}{a  - 1}

EXPLANATION

The given function is

f(x) =  \frac{3 + x}{x - 3}

To find f(a+2) we substitute x=a+2.

That is wherever we see 'x' we replace it with 'a+2'

f(a + 2) =  \frac{3 + a + 2}{a + 2 - 3}

We simplify further to obtain:

f(a + 2) =  \frac{a + 5}{a  - 1}

This is defined for

a \ne1

5 0
4 years ago
The price of a dirt bike dropped from $872 to $600
yuradex [85]

Answer:

thats sick

Step-by-step explanation:

go kop that

5 0
4 years ago
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