Can you explain these questions and how you got the answers?

B

A

Step-by-step explanation:

To find f'(x), take the derivative within each interval.  Clearly, it's 5 in the first one, 2x in the second one, and 4 in the third one.  But we need to determine if the derivative exists at the ends of each interval.

For a derivative to exist at a point x=a, the function must be continuous (f(a⁻) = f(a⁺)), and smooth (f'(a⁻) = f'(a⁺)).

Let's look at x=4.  On the left side:

f(4⁻) = 14, f'(4⁻) = 5

On the right side:

f(4⁺) = 14, f'(4⁺) = 8

So the function is continuous, but not smooth.  Therefore, the derivative doesn't exist there.

Let's try again with x=6.  On the left side:

f(6) = 34, f'(6) = 12

On the right side:

f(6⁺) = 34, f'(6⁺) = 4

Again, the function is continuous, but not smooth, so the derivative doesn't exist there either.

Therefore, f'(x) is:

When we graph f(x) = (4x² − 1) / (x² − 9), we see f(x) has a horizontal tangent line at x=0.

desmos.com/calculator/6y3r1ju7jx

{x = -1/2 ,   y = -1

step-by-step explanation:

solve the following system:

{2 x + y = -2

{10 x - 3 y = -2

in the first equation, look to solve for y:

{2 x + y = -2

{10 x - 3 y = -2

subtract 2 x from both sides:

{y = -2 x - 2

{10 x - 3 y = -2

substitute y = -2 x - 2 into the second equation:

{y = -2 x - 2

{10 x - 3 (-2 x - 2) = -2

10 x - 3 (-2 x - 2) = (6 x + 6) + 10 x = 16 x + 6:

{y = -2 x - 2

{16 x + 6 = -2

in the second equation, look to solve for x:

{y = -2 x - 2

{16 x + 6 = -2

subtract 6 from both sides:

{y = -2 x - 2

{16 x = -8

divide both sides by 16:

{y = -2 x - 2

{x = -1/2

substitute x = -1/2 into the first equation:

{y = -1

{x = -1/2

collect results in alphabetical order:

answer:   {x = -1/2 ,   y = -1

step-by-step explanation:

i found the life tables for the year 2012. 

life table of males at age 30. probability of dying between ages x and x+1

total population:  

29–30 . . . . . . . . . . . . 0.001030

30–31 . . . . . . . . . . . . 0.001061

males:  

29–30 . . . . . . . . . . . . 0.001420

30–31 . . . . . . . . . . . . 0.001448

white population

29–30 . . . . . . . . . . . . 0.001005

30–31 . . . . . . . . . . . . 0.001037

white males

29–30 . . . . . . . . . . . . 0.001372

30–31 . . . . . . . . . . . . 0.001399

non hispanic white population

29–30 . . . . . . . . . . . . 0.001083

30–31 . . . . . . . . . . . . 0.001116

non hispanic white males

29–30 . . . . . . . . . . . . 0.001477

30–31 . . . . . . . . . . . . 0.001503

jacob, non hispanic white male. 30 years old.

probability that jacob lives to be 30 years old. 

since the given probabilities are probabilities of dying, i think we simply subtract it from 1 to get the probability of living. we will use the probability of non hispanic white male.

                                          probability of living to be 30 y.o.

29 - 30 : 1 - 0.001477 = 0.998523

30 - 31 : 1 - 0.001503 = 0.998497