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Arte-miy333 [17]
3 years ago
13

Read the problem.

Mathematics
1 answer:
Brut [27]3 years ago
4 0

Answer:

I think that Sarah sent 70 percent of the text messages , so you have to shade 70 of them .

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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
A circle with circumference 20 has an arc with a 72 degree central angle. What is the length of the arc?
drek231 [11]

Answer:

4

Step-by-step explanation:

circumference = 2πr

or, 20 = 2π r

or, π r = 10

or, r= 10/π

72 ° = 2π/5

length of arc = r∅ = 10/π × 2π/5

= 4

5 0
3 years ago
Evaluate using the trigonometric ratios Tan 60° + Cos 30° sin 60° + Cot 30° /CSC 30° Sec 30°<br>​
algol13

Answer:

(24 + 3√3) / 16.

Step-by-step explanation:

Tan 60° + Cos 30° sin 60° + Cot 30 /  csc 30 sec 30

= √3 + √3/2 * √3/2 + √3 /   2 * 2/√3

= 2√3 + 3/4 / 4/√3

= (8√3 + 3)/ 4 / √3

= 24 + 3√3 / 16

5 0
2 years ago
What is the area of a rectangle with side lengths 2/5 feet and 4/6 feet?
kotegsom [21]

Answer:

\frac{4}{15}  (4/15)

Step-by-step explanation:

\frac{4}{6}=\frac{2}{3}

\frac{2}{5}*\frac{2}{3};

1- Multiply the numerators:

2*2=4

2- Multiply the denominators:

5*3=15

3- Thus:

\frac{2}{5}*\frac{2}{3}= \frac{4}{15}

Hope this helps ;)

4 0
3 years ago
What is 4 x 2? Help.
OverLord2011 [107]

Answer:

8

Step-by-step explanation:

5 0
3 years ago
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