Response:
45.54m
Step-by-step breakdown:
See the provided figure
Scarlett's height is represented as AB = 1.65 meters
She stands 90 meters away from the dam, so BE = AC = 90 m
Scarlett measures the angle of elevation to the dam's summit at 26º, which is ∠DBE = 26°
Height of the Dam = DC = EC+DE
AB = EC = 1.65
In triangle BDE
Height of the Dam = DC = 1.65 + 43.89 = 45.54 m
Therefore, the dam's height is 45.54 m.
RT equals RS plus ST
8x - 43 = 2x - 4 + 4x - 1
Simplify: 8x - 43 = 6x - 5
Bring variables to one side: 8x - 6x = -5 + 43
2x = 38
Divide both sides by 2:
x = 19
Calculate QS as twice RS:
QS = 2 * RS = 2 * (2x - 4) = 2 * (2*19 - 4) = 2 * 34 = 68
Answer:
The function decreases across all real numbers x where x < 1.5.
Step-by-step explanation:
We have
This represents an upward-opening vertical parabola
The vertex denotes a minimum point
The vertex is located at (1.5,-6.25)
We know that
The function is decreasing within the interval ----> (-∞, 1.5) x < 1.5
This means----> the function is continuously decreasing for all real x values under 1.5
Conversely, the function is increasing within the interval ----> (1.5, ∞) x> 1.5
Thus, for all real x values greater than 1.5, the function is increasing
Check the attached figure for further clarification
Therefore
The true statement is
The function decreases across all real numbers x where x < 1.5.
Given that,
Julia completes a 20-mile bike ride in 1.2 hours.
The distance Julia covers is 20 miles and her time taken is 1.2 hours.
Therefore, Julia's speed = 
= 16.67 mph
Katie finishes the same 20-mile ride in 1.6 hours.
Katie’s distance is 20 miles and her time is 1.6 hours.
Hence, Katie's speed = 
= 12.5 mph
To determine how much faster Julia rides compared to Katie, subtract Katie’s speed from Julia’s speed.
Thus, 16.67 mph minus 12.5 mph equals 4.17 mph, approximately 4.2 mph.
Consequently, Julia cycles 4.2 mph faster than Katie.
To formulate the system, it's necessary to consider the slope of each line along with at least one point from each line. The two lines will connect each plane's location to their destination airport. It's important to note that the airport's coordinates represent the intersection of these two lines, corresponding to the solution of the system. First, the slope of the line from airplane one to the airport is: m = 2; this can be observed by plotting the two points. From airplane 1's location, the rise is 8 units while the run is 4 units to reach the airport, making the slope 8 divided by 4 = 2. We then insert the slope and point (2,4) into the point-slope form: y - 4 = 2(x - 4), which can be rearranged to standard form 2x - y = 0. For airplane two, the slope to the airport is obtained by observing the vertical decrease of 3 and a horizontal increase of 9 as we move from the airport to airplane 2. We then substitute the slope and the point (15,9) into the point-slope form: y - 9 = -1/3(x - 15), which can be rearranged to the standard form: x + 3y = 42. Consequently, the system of equations is: 2x - y = 0 and x + 3y = 42. Multiplying the first equation by 3 produces a system of: 6x - 3y = 0 and x + 3y = 42. Adding these equations results in the equation 7x = 42. Thus, x = 6, and by substituting this value back into 2x - y = 0, we determine y = 12. Thus, we demonstrate that the airport's coordinates do indeed comprise the solution to our system.
