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WITCHER [35]
3 years ago
13

F(x)=x^2+10 Over which interval does fff have a positive average rate of change?

Mathematics
2 answers:
ohaa [14]3 years ago
6 0

Answer:

(0, inf)

Step-by-step explanation:

The average rate of change of a function is related to it's first derivative. When the first derivative is positive, the average rate of change is positive, which means that the function is crescent.

Now, when the first derivative is negative, the average rate of change will be negative too, and the function is going to be decrescent.

In your function.

We have: f'(x) = 2x

2x > 0 when x > 0. So when x > 0 the average rate of change of your function is positive, and it's values increases as the time increases. When x < 0, the average rate of change is negative, so, as the time increases, the values of f decrease.

You can use a graphic tool to plot f and visualize this better.

Allushta [10]3 years ago
6 0

Answer:

The answer is  [-1,2]

Hope it helps

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jonny [76]

Answer:

Yes Maggie has enough money to purchase the dress and hair accessory.

Step-by-step explanation:

Dress = $35.75

Hair accessory = $2.80

Total amount of money = $40

Does her have enough money?

Total Money in pocket - (Dress cost + hair accessory)

 $40.00 - ($35.75 + 2.80)

  $40.00 - ($36.00 + $3.00)     Round to the nearest dollar

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Yes

8 0
2 years ago
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Step-by-step explanation:

\frac{\sin \:4A}{\cos \:2A}  \times \frac{1 - \cos \:2A}{1 - \cos\:4A}   = \tan \:A \\  \\ LHS =  \frac{\sin \:4A}{\cos \:2A}  \times \frac{1 - \cos \:2A}{1 - \cos\:4A}   \\  \\  =  \frac{2\sin \:2A.\cos \:2A}{\cos \:2A}   \times  \frac{1 - (2 { \cos}^{2}A - 1) }{1 - (2 { \cos}^{2}2A - 1) } \\  \\  = 2\sin \:2A   \times  \frac{1 - 2 { \cos}^{2}A  +  1}{1 - 2 { \cos}^{2}2A  + 1 } \\  \\  = 2\sin \:2A   \times  \frac{2- 2 { \cos}^{2}A  }{2 - 2 { \cos}^{2}2A   } \\  \\   = 2\sin \:2A   \times  \frac{2(1 - { \cos}^{2}A)  }{2 (1-  { \cos}^{2}2A)   } \\  \\   = 2\sin \:2A   \times  \frac{1 - { \cos}^{2}A}{1-  { \cos}^{2}2A   } \\  \\     = 2\sin \:2A   \times  \frac{ { \sin}^{2}A}{{ \sin}^{2}2A   } \\  \\    = 2  \times  \frac{ { \sin}^{2}A}{{ \sin}2A   } \\  \\   = 2  \times  \frac{ { \sin}^{2}A}{{ 2\sin}A. \cos \:   A } \\  \\   = \frac{ { \sin}A}{ \cos \:   A }  \\  \\  = tan \: A \\  \\  = RHS \\

7 0
3 years ago
What is the equation of the circle with center (-3,1) that passes through the point (-5, 3)?
Natali5045456 [20]

Answer:

Option B) (x + 3)^2 + (y – 1)^2 = 8 is the correct answer.

Step-by-step explanation:

The equation of a circle with center (h,k) and radius r is given by:

(x-h)^2 + (y-k)^2 = r^2

Given

Center = (h,k) = (-3,1)

=> h = -3

=> k = 1

The distance between the center of circle and the point through which the circle passes will be the radius.

The distance formula is given by:

r = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2

Given

(x_1,y_1) = (-3,1)\\(x_2,y_2) = (-5,3)

Putting the values in the formula

r = \sqrt{(-5+3)^2+(3-1)^2}\\r = \sqrt{(-2)^2+(2)^2}\\r = \sqrt{4+4}\\r = \sqrt{8}

Putting the values of h,k and r in general form of equation

\{x-(-3)}^2\} +(y-1)^2 = (\sqrt{8})^2\\(x+3)^2+(y-1)^2 = 8

Hence,

Option B) (x + 3)^2 + (y – 1)^2 = 8 is the correct answer.

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