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2 years ago

Let f(x)=cos(x)x^-2 F’(x)=

1 answer:

8 0

$f(x)=cos(x)x^{-2}\implies f(x)=\cfrac{cos(x)}{x^2}\implies \cfrac{df}{dx}=\stackrel{\textit{quotient rule}}{\cfrac{-sin(x)\cdot x^2-cos(x)\cdot 2x}{(x^2)^2}} \\\\\\ \cfrac{df}{dx}=\cfrac{-x[x\cdot sin(x)+2cos(x)]}{x^4}\implies \cfrac{df}{dx}=\cfrac{-[xsin(x)+2cos(x)]}{x^3}$

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Please help me with question 8 and part 1

larisa [96]

Answer:

m∠B = 95°

Scalene

Step-by-step explanation:

by definition, internal angles in a triangle must add up to 180°

in this case,

∠A + ∠B + ∠C = 180°

we are given ∠C = 55° and ∠A = 30°

30° + ∠B + 55° = 180°

∠B = 180 - 55 - 30 = 95° (Answer)

We can see that all 3 angles of the triangle are different. And hence we can conclude that all 3 sides of the triangle are different length (i.e it is not equilateral or isosceles). Hence it is a scalene triangle.

5 0

3 years ago

Solve the equation.<br> 2/5n+1/10=1/2 (n+4)

viktelen [127]

The answer would be (n=19)

6 0

3 years ago

A die is rolled two separate times. What is the probability that the first roll is a 4 and then the second roll is a 2? *

padilas [110]

Answer:

So we know that rolling a fair six-sided die twice would mean the total possible outcomes would be 36, and rolling the 4 on the first roll and a 2 on the second roll would be 2/36 or 1/18

Step-by-step explanation:

(1/6)*(1/6)=(2/36)=(1/18)

6 0

3 years ago

There are 12 appetizers; 4 are soups, 6 contain meat, and 2 do not. In how

Ahat [919]

72 different orders, each soup can have a different meat and other* so take 4* 6* 3 because that is how many different combanations you can do

7 0

3 years ago

Read 2 more answers

Determine the location and values of the absolute maximum and absolute minimum for given function : f(x)=(‐x+2)4,where 0<×&lt

brilliants [131]

Answer:

Where 0 < x < 3

The location of the local minimum, is (2, 0)

The location of the local maximum is at (0, 16)

Step-by-step explanation:

The given function is f(x) = (x + 2)⁴

The range of the minimum = 0 < x < 3

At a local minimum/maximum values, we have;

$f'(x) = \dfrac{(-x + 2)^4}{dx} = -4 \cdot (-x + 2)^3 = 0$

∴ (-x + 2)³ = 0

x = 2

$f''(x) = \dfrac{ -4 \cdot (-x + 2)^3}{dx} = -12 \cdot (-x + 2)^2$

When x = 2, f''(2) = -12×(-2 + 2)² = 0 which gives a local minimum at x = 2

We have, f(2) = (-2 + 2)⁴ = 0

The location of the local minimum, is (2, 0)

Given that the minimum of the function is at x = 2, and the function is (-x + 2)⁴, the absolute local maximum will be at the maximum value of (-x + 2) for 0 < x < 3

When x = 0, -x + 2 = 0 + 2 = 2

Similarly, we have;

-x + 2 = 1, when x = 1

-x + 2 = 0, when x = 2

-x + 2 = -1, when x = 3

Therefore, the maximum value of -x + 2, is at x = 0 and the maximum value of the function where 0 < x < 3, is (0 + 2)⁴ = 16

The location of the local maximum is at (0, 16).

5 0

3 years ago