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Lady_Fox [76]
3 years ago
7

If the area is 2289.06 whats the radius

Mathematics
1 answer:
solniwko [45]3 years ago
3 0
I'm assuming you are referring to the radius of a square

The first thing we have to do is divide by pi, because it's the last thing you do when calculating the area.

2289.06/3.14
729

Now, we have to take the square root of this number because the radius in the area formula is squared.

729^1/2
27

So, the radius is 27.
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Answer:

21+3x

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3(7+x)

3*7+3*X

21+3x

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Frosty Whip, Inc. budgets 65 percent of its total expenses for marketing. Total expenses for the company this year are $500,000.
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During a sale at a bookstore, a customer buys one book at full price. The customer is given a 25 percent discount on a second bo
RoseWind [281]
Amount paid for the first book is $17.

Amount paid for the second book is 0.75 x 8 = $6

Total amount paid = $17 + $6 = $23

Total full cost of the two books = $17 + $8 = $25

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% reduction = \frac{2}{25} \times100\%=8\%

Therefore, the <span>percent of the total cost of the two books reduced during the sale</span> is 8%
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What is absolute extrema of cube root of x on I=[-3,8]
hichkok12 [17]
<span>These are points where f ' = 0. Use the quiotent rule to find f '. 

f ' (x) = [(x^3+2)(1) - (x)(3x^2)] / (x^3+2)^2 
f ' (x) = (2 - 2x^3) / (x^3 + 2)^2 

Set f ' (x) = 0 and solve for x. 

f ' (x) = 0 = (2-2x^3) / (x^3+2)^2 

Multiply both sides by (x^3+2)^2 

(x^3+2)^2 * 0 = (x^3+2)^2 * [(2-2x^3)/(x^3+2)^2] 
0 = 2 - 2x^3 

Add 2x^3 to both sides 

2x^3 + 0 = 2x^3 + 2 - 2x^3 
2x^3 = 2 

Divide both sides by 2 

2x^3 / 2 = 2 / 2 
x^3 = 1 

Take cube roots of both sides 

cube root (x^3) = cube root (1) 
x = 1. This is our critical point 

2) Points where f ' does not exist. 

We know f ' (x) = (2-2x^3) / (x^3+2)^2 

You cannot divide by 0 ever so f ' does not exist where the denominator equals 0 

(x^3 + 2)^2 = 0. Take square roots of both sides 
sqrt((x^3+2)^2) = sqrt(0) 
x^3 + 2 = 0. Add -2 to both sides. 
-2 + x^3 + 2 = -2 + 0 
x^3 = -2. Take cube roots of both sides. 
cube root (x^3) = cube root (-2) 
x = cube root (-2). This is where f ' doesnt exist. However, it is not in our interval [0,2]. Thus, we can ignore this point. 

3) End points of the domain. 

The domain was clearly stated as [0, 2]. The end points are 0 and 2. 

Therefore, our only options are: 0, 1, 2. 

Check the intervals 

[0, 1] and [1, 2]. Pick an x value in each interval and determine its sign. 

In [0, 1]. Check 1/2. f ' (1/2) = (7/4) / (17/8)^2 which is positive. 

In [1, 2]. Check 3/2. f ' (3/2) = (-19/4) / (43/8)^2 which is negative. 

Therefore, f is increasing on [0, 1] and decreasing on [1, 2] and 1 is a local maximum. 

f (0) = 0 
f (1) = 1/3 
f (2) = 1/5 

Therefore, 0 is a local and absoulte minimum. 1 is a local and absolute
maximum. Finally, 2 is a local minimum. </span><span>Thunderclan89</span>
3 0
4 years ago
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