Harmony earns a \$42{,}000$42,000dollar sign, 42, comma, 000 salary in the first year of her career. Each year, she gets a 4\%4%

Answer:

The hourly rate for lectures is $7.33

Step-by-step explanation:

* Let's break down how to tackle the problem.

- For the level 3 course, examination hours are priced at double that of workshop hours.

- Workshop hours cost twice the rate of lecture hours.

- The total includes examination, workshop, and lecture hours.

- Examination lasts 3 hours, workshops 24 hours, and lectures 12 hours.

* Let’s denote the cost of lecture hours as $x per hour.

∴ The lectures cost $x per hour.

∵ Workshop charge is twice that of lectures

∴ Workshop hours cost 2(x) = 2x per hour.

∵ Examination fees are double that of workshop hours

∵ The workshop cost is 2x

∴ Examination fees are 2(2x) = 4x per hour.

- Combining costs for level 3 gives us the total of lecture, workshop, and examination hours.

∵ 12 hours for lectures

∵ 24 hours for workshops

∵ 3 hours for examinations

∵ Thus the total cost for level 3 = 12(x) + 24(2x) + 3(4x).

∴ Total cost for level 3 = 12x + 48x + 12x.

∵ Therefore, total cost = $528.

∴ 12x + 48x + 12x = 528.

∴ 72x = 528; hence we divide both sides by 72.

∴ x = 7.33.

∵ x represents the cost of lecture hours per hour.

∴ Therefore, the hourly price for lectures is $7.33.

The fox’s speed needs to exceed that of the rabbit to successfully catch up. Thus, the required speed for the fox is expressed as 35/t + 40.

The distance separating the rabbit from the fox is 35 feet.

The speed of the rabbit is 40 feet per second.

Fox's speed = F

To calculate a precise speed for the fox, a specific time for the pursuit must be provided.

Let’s denote the time as t.

Recall that:

Speed = distance / time

Distance = speed × time

The distance the rabbit moves away from the fox after time t will be:

35 + (40 × t) = 35 + 40t

The distance traveled by the fox after time t is: fox speed × t = F × t.

To catch the rabbit, both the fox and rabbit must cover the same distance:

Rabbit's distance at time t equals Fox's distance at that time.

35 + 40t = Ft

To determine F, the fox's speed:

Dividing both sides by t yields:

(35 + 40t) / t = Ft/t

35/t + 40 = F

<pThus, the speed of the fox should be: 35/t + 40

To find out the exact speed of the fox, a specific time value must be provided.

Learn more:

Density can be defined as:

Where:

m is mass

V stands for volume

To isolate the mass, we derive:

Volume can be expressed as:

We then use the following conversion:

After applying the conversion, we arrive at:

Additionally, we have these conversions:

When we apply these conversions for density, we conclude:

Finally, the mass of the water required is:

Answer:

A total of 1190 kilograms of water is needed to fill the waterbed.

Starting with the equation 4(3b + 2)² = 64, if we divide both sides by 4 we obtain (3b + 2)² = 16. By taking the square root of both sides, we derive two cases: (3b + 2) = 4 and (3b + 2) = -4. Solving each equation for b yields: 3b = 2 or 3b = -6, leading to b values of 2/3 and -2. Ultimately, the results specify that b = 2/3 and b = -2.

Answer:

a) y = 7.74*x + 7.5

b)  y = 1.148*x + 6.036

Step-by-step explanation:

Given:

                                  f(x) = 6 - 10*x^2

                                  g(x) = 8 - (x-2)^2

Find:

(a) The equation of the line with the LARGEST slope that is tangent to both graphs is

(b) The equation of the second line tangent to both graphs is:

Solution:

- First, let's calculate the derivatives for the two functions provided:

                                f'(x) = -20*x

                                g'(x) = -2*(x-2)

- Given that the derivatives of both functions are dependent on the x-value, we will pick a common point x_o for both f(x) and g(x). This point is ( x_o, g(x_o)). Thus,

                                g'(x_o) = -2*(x_o - 2)

- Next, we will determine the slope of a line that is tangent to both graphs at point (x_o, g(x_o) ) on g(x) and at ( x, f(x) ) on f(x):

                                m = (g(x_o) - f(x)) / (x_o - x)

                                m = (8 - (x_o-2)^2 - 6 + 10*x^2) / (x_o - x)

                                m = (8 - (x_o^2 - 4*x_o + 4) - 6 + 10*x^2)/(x_o - x)

                                m = ( 8 - x_o^2 + 4*x_o -4 -6 +10*x^2) /(x_o - x)

                                m = ( -2 - x_o^2 + 4*x_o + 10*x^2) /(x_o - x)

- Now we need to set the slope from our equation equal to the derivatives we calculated earlier for each function:

                                m = f'(x) = g'(x_o)

- We will work through the first expression:

                                m = f'(x)

                                ( -2 - x_o^2 + 4*x_o + 10*x^2) /(x_o - x) = -20*x

Eq 1.                          (-2 - x_o^2 + 4*x_o + 10*x^2) = -20*x*x_o + 20*x^2

And,

                              m = g'(x_o)

                              ( -2 - x_o^2 + 4*x_o + 10*x^2) /(x_o - x) = -20*x

                              -2 - x_o^2 + 4*x_o + 10*x^2 = -2(x_o - 2)(x_o - x)

Eq 2                       -2 - x_o^2 + 4*x_o+ 10*x^2 = -2(x_o^2 - x_o*(x + 2) + 2*x)

- Now we can subtract the two equations (Eq 1 - Eq 2):

                              -20*x*x_o + 20*x^2 + 2*x_o^2 - 2*x_o*(x + 2) + 4*x = 0

                              -22*x*x_o + 20*x^2 + 2*x_o^2 - 4*x_o + 4*x = 0

- Rearranging gives us:       20*x^2 - 20*x*x_o - 2*x*x_o + 2*x_o^2 - 4*x_o + 4*x = 0

                              20*x*(x - x_o) - 2*x_o*(x - x_o) + 4*(x - x_o) = 0

                               (x - x_o)(20*x - 2*x_o + 4) = 0  

                               x = x_o,     x_o = 10x + 2    

- For x_o = 10x + 2,

                               (g(10*x + 2) - f(x))/(10*x + 2 - x) = -20*x

                                (8 - 100*x^2 - 6 + 10*x^2)/(9*x + 2) = -20*x

                                (-90*x^2 + 2) = -180*x^2 - 40*x

                                90*x^2 + 40*x + 2 = 0  

- Now solving the above quadratic equation:

                                 x = -0.0574, -0.387      

- The maximum slope occurs at x = -0.387, with the line’s equation being:

                                  y - 4.502 = -20*(-0.387)*(x + 0.387)

                                  y = 7.74*x + 7.5          

- The second tangent line is:

                                  y - 5.97 = 1.148*(x + 0.0574)

                                  y = 1.148*x + 6.036