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soldier1979 [14.2K]
3 years ago
11

HELP YOU WILL GET BRAINLIST

Mathematics
1 answer:
aksik [14]3 years ago
4 0

Answer:

it is the blue box (not teal)

64 x 7/4 is less than 64 because 7/4 > 1

Hope this helps :)

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I WILL CHOOSE BRAINLIEST!!!
sergejj [24]

Answer:

13.8

Step-by-step explanation:

the required constant can be calculated as:

c=\frac{FinalWages-InitialWages}{FinalTime-InitialTime} ;

according to the formula above:

c=\frac{82.8-27.6}{6-2} =13.8.

5 0
3 years ago
Why does the series not exist?
Ber [7]

Step-by-step explanation:

(7n − 5) / (2n + 5) approaches 7/2 as n approaches infinity.  However, this is an alternating series.  So the function is fluctuating between -7/2 and +7/2.  So the limit does not exist.

5 0
3 years ago
What is MJ’s slugging average?
Ivenika [448]
In 127 games for the Barons, MJ managed just three home runs, 17 doubles, and one triple. All told, it amounted to a . 266 slugging percentage, which is puny by any measur
5 0
3 years ago
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
How can I solve this equation 5x^2-2x-6=-2x^2+9x
wel
Add 2x^2 to both sides
7x^2-2x-6=9x

Subtract 9x from both sides
7x^2-11x-6=0

Factor by using slip and slide
x^2-11x-42
(x-14)(x+3)
(x-14/7)(x+3/7)
(x-2)(7x+3)

Then use the Zero Product Property to find roots
x-2=0
x=2

7x+3=0
7x=-3
x=-3/7

Final answer: x=-3/7, x=2
3 0
3 years ago
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