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Dima020 [189]
3 years ago
8

Represent each of the following variables in the appropriate letters.Next determine wether the variable has a fixed value or a c

hanged value.
a)The score Sarah got in a Mathematics test.
b)The distance between the Zen house and the school.
c)Temperature on the summit of Mount Everest in a day.​
Mathematics
1 answer:
Alex777 [14]3 years ago
6 0

Step-by-step explanation:

a) Let 'm' represents score in a Mathematics test. The variable has a fixed value because the marks Sarah obtained in her test will not change

b) Let 'd' represent the distance between the Zen house and the school. The variable has a fixed value because by using the same route, the distance will always remain constant.

c) Let 't' represent the temperature on the summit of Mount Everest. The variable has a changed value because the temperature may alter depending on the time of day and also climatic conditions which will affect the temperature.

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y=-3/2x+4

Step-by-step explanation:

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2 years ago
Read 2 more answers
Find the 2nd Derivative:<br> f(x) = 3x⁴ + 2x² - 8x + 4
ad-work [718]

Answer:

f''(x)=36x^2+4

Step-by-step explanation:

Let's start by finding the first derivative of f(x)= 3x^4+2x^2-8x+4. We can do so by using the power rule for derivatives.

The power rule states that:

  • \frac{d}{dx} (x^n) = n \times x^n^-^1

This means that if you are taking the derivative of a function with powers, you can bring the power down and multiply it with the coefficient, then reduce the power by 1.

Another rule that we need to note is that the derivative of a constant is 0.

Let's apply the power rule to the function f(x).

  • \frac{d}{dx} (3x^4+2x^2-8x+4)

Bring the exponent down and multiply it with the coefficient. Then, reduce the power by 1.

  • \frac{d}{dx} (3x^4+2x^2-8x+4) = ((4)3x^4^-^1+(2)2x^2^-^1-(1)8x^1^-^1+(0)4)

Simplify the equation.

  • \frac{d}{dx} (3x^4+2x^2-8x+4) = (12x^3+4x^1-8x^0+0)
  • \frac{d}{dx} (3x^4+2x^2-8x+4) = (12x^3+4x-8(1)+0)
  • \frac{d}{dx} (3x^4+2x^2-8x+4) = (12x^3+4x-8)
  • f'(x)=12x^3+4x-8

Now, this is only the first derivative of the function f(x). Let's find the second derivative by applying the power rule once again, but this time to the first derivative, f'(x).

  • \frac{d}{d} (f'x) = \frac{d}{dx} (12x^3+4x-8)
  • \frac{d}{dx} (12x^3+4x-8) = ((3)12x^3^-^1 + (1)4x^1^-^1 - (0)8)

Simplify the equation.

  • \frac{d}{dx} (12x^3+4x-8) = (36x^2 + 4x^0 - 0)
  • \frac{d}{dx} (12x^3+4x-8) = (36x^2 + 4(1) - 0)
  • \frac{d}{dx} (12x^3+4x-8) = (36x^2 + 4 )

Therefore, this is the 2nd derivative of the function f(x).

We can say that: f''(x)=36x^2+4

6 0
2 years ago
Read 2 more answers
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