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dexar [7]
3 years ago
8

FIVE STARS AND BRAINLIEST TO CORRECT ANSWER

Mathematics
1 answer:
timama [110]3 years ago
7 0

Given, f(x) = 4/x

We have to find the derivative of it.

First we have to use property of negative exponents. We know that, 1/a = a⁻¹.

So here, f(x) = 4/x = 4x⁻¹

f'(x) = 4×(-1)×(x)⁽⁻¹⁻¹⁾

= 4×(-1)×(x)⁻²

= (-4)(x)⁻²

= -4/x²

So we have got derivative of f(x) which is f'(x) = -4/x²

Now we have to find the derivative at x = 2. To find the value at x =2, we have to plug in 2 in the place of x.

f'(2) = -4/(2)² = -4/4 = -1

We have got the required answer derivative of f(x) at x =2 is -1

So option B is correct.

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erastova [34]

Answer:

Q3: x = 4, y = 4, z = 4

Q4: x = 6, y = 0, z = -4

Step-by-step explanation:

Question 3: Simultaneous equations requires us to solve for x, y and z.

Since all three equations have a z in them, I will first solve for z.

Substitute in the first and third equation into the second equation.

First equation: x = 5z - 16

Second Equation: -4x + 4y - 5z = -20

Third equation: y = -z + 8

Substituting in x = 5z - 16 and y = -z + 8 for the x and y in the second equation.

-4(5z - 16) + 4(-z + 8) - 5z = -20

Expand

-20z + 64   - 4z + 32   - 5z = -20

Simplify and solve for z by putting all the numbers on one side and all the z's on the other side of the equals

-20z - 4z - 5z = -20 - 32 - 64

-29z = -116

z = -116/-29

z = 4

Substitute in this z value into the first and last equation and then solve for x and y

x = 5z - 16

x = 5(4) - 16

x = 20 - 16

x = 4

And

y = -z + 8

y = -(4) + 8

y = 4 (Its just a coincidence that they all equal to 4, I promise)

Question 5: A little bit harder of a question. Since the first and second equation both only have y and z, we can solve it using the elimination method.

Rearrange them so that the letters are on one side and numbers on the other side.

First equation: y + 6z = -24

Second equation: z + 2y = -4

I will choose to eliminate the y (You can choose either or)

Multiply the first equation by 2

2(y + 6z = -24)

2y + 12z = -48

Now that 2y is in both equations, we can minus one equation from the other to eliminate the y (I will minus the second from the first)

First Eq: 2y + 12z = -48

Second Eq: z + 2y = -4

2y - 2y = 0y

12z - z = 11z

-48 - (-4) = -44

Type these answers into a new equation

0y + 11z = - 44

Since y is 0, ignore it. Solve for z

11z = -44

z = -44/11

z = - 4

Substitute our z into either the first or second equation and solve for y (It doesnt matter which one you choose, I just did the second equation)

z + 2y = -4

(-4) + 2y = -4

2y = -4 + 4

2y = 0

y = 0

Substitute in our y and z values into the third equation and solve for x

-6x - 6y - 6z = -12

-6x - 6(0) - 6(-4) = -12

-6x - 0 + 24 = -12

-6x = -12 - 24

-6x = -36

x = -36/-6

x = 6

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