Which are the solutions of x2 = –11x + 4? StartFraction negative 11 minus StartRoot 137 EndRoot Over 2 EndFraction comma StartFr

Average fall rate =

Detailed explanation:

Given that the position of the object is expressed as a function of time, we can compute the average velocity of the object during the first 3 seconds by determining its position at time 0 and at 3 seconds, finding the displacement over that period, and then dividing this distance by the elapsed time (3 seconds). This approach follows the concept that velocity equals distance covered divided by time spent:

Average rate of fall = (h(3) - h(0))/ 3

Average rate of fall =

The statements labeled 1 and 4 are accurate. To easily see the center and radius of the circle, we can modify the given equation to fit its standard format. Once in standard form, we can contrast it with the standard equation to find the circle's center and radius. The coordinates for the center are determined to be (1,0) and the radius is represented by . With this understanding, we can evaluate each statement. 1. The circle's radius is 3 units—this is true. 2. The circle's center is located on the y-axis—this is incorrect, as the center at (1,0) indicates it is on the x-axis. 3. The standard equation is (x - 1)² + y² = 3—this is false; the correct equation is (x - 1)² + y² = 9. 4. The circle's radius matches that of the circle with the equation x² + y² = 9—this statement is correct, as both radii equal 3.

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

Answer:

• Boat speed: 17.25 mph
• Current: 2.5 mph

Step-by-step explanation:

When going upstream, the current's speed (c) reduces the boat's speed (b), resulting in a net speed. Conversely, going downstream, the current's speed adds to the boat's speed. We can use the relation...

speed = distance/time

b - c = 29.5/2 = 14.75..... miles/hour

b + c = 59.25/3 = 19.75....miles/hour

By subtracting the first equation from the second, we arrive at...

(b+c) - (b-c) = 19.75 - 14.75

2c = 5

c = 2.5

Then we can use any of the equations to calculate b:

b = 14.75 + c = 17.25

The boat's speed in still water amounts to 17.25 miles per hour, while the current's speed is 2.5 miles per hour.

d Step-by-step explanation: when f(x)=5......It can also be expressed as f(x)=1/125(625x). This is applicable to any whole number over one, with 1(625x)=625x. Therefore, since 125(1)=125, we find (625/125)x=5x.