Taking the derivative of 7 times secant of x^3:
We take out 7 as a constant focus on secant (x^3)
To take the derivative, we use the chain rule, taking the derivative of the inside, bringing it out, and then the derivative of the original function. For example:
The derivative of x^3 is 3x^2, and the derivative of secant is tan(x) and sec(x).
Knowing this: secant (x^3) becomes tan(x^3) * sec(x^3) * 3x^2. We transform tan(x^3) into sin(x^3)/cos(x^3) since tan(x) = sin(x)/cos(x). Then secant(x^3) becomes 1/cos(x^3) since the secant is the reciprocal of the cosine.

We then multiply everything together to simplify:

sin(x^3) * 3x^2/ cos(x^3) * cos(x^3) becomes

3x^2 * sin(x^3)/(cos(x^3))^2

and multiplying the constant 7 from the beginning:

7 * 3x^2 = 21x^2, so...

our derivative is 21x^2 * sin(x^3)/(cos(x^3))^2

6 0

3 years ago

Solve the equation 2/3-4x+7/2=-9x+5/6

VMariaS [17]

Answer:

For this equation the value of x is - 2/3

Step-by-step explanation:

1. Resolving the equation 2/3-4x+7/2=-9x+5/6

-4x + 9x = 5/6 - 2/3 - 7/2 (Putting the all the x values on the left)

5x = (5/6 - 4/6 -21/6)

5x = -20/6

x = - 20/6 /5 (Dividing by 5 at both sides)

x = -20/6 * 1/5

x = -4/6 = - 2/3 (Simplifying)

2. Proof of replacing x by -2/3

2/3 - 4 (-2/3) + 7/2 = -9 (-2/3) + 5/6

2/3 + 8/3 + 7/2 = 18/3 + 5/6

10/3 + 7/2 = 6 + 5/6

20 + 21 = 36 + 5 (Multiplying by 6 at both sides)

41 = 41

It means the value of -2/3 for x is correct

Note: Same answer than 13866851

4 0

3 years ago

Follow the steps above and find c, the total of payments of the payments, and the monthly payment. Choose the right answers. Jan

Alex_Xolod [135]

Whats the answers people come on

7 0

3 years ago

The number of chocolate chips in a bag of chocolate chip cookies is approximately normally distributed with mean of 1262 and a s

Andrew [12]

Answer:

a) 1186

b) Between 1031 and 1493.

c) 160

Step-by-step explanation:

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Normally distributed with mean of 1262 and a standard deviation of 118.

This means that $\mu = 1262, \sigma = 118$