The
correct illustration is provided.
Explanation:
Utilizing a tool like Geogebra, commence by creating a line segment. Label the endpoints as C and D.
Next, draw the perpendicular bisector of the segment and denote the intersection with CD as B, then introduce a point A above this line.
Measure the distance from C to B and from B to D. Both distances will be equal.
Measure the length from A to B. If this distance is not equal to that from C to B, adjust A along line AB until the distances match.
With a compass and straightedge:
First, create segment CD and ensure the endpoints are labeled.
Adjust your compass to slightly more than half the distance between C and D. With it set at C, draw an arc above CD.
Using the same compass setting at D, draw another arc to intersect your first arc above CD. Mark the intersection as E.
Connect E to CD with a straightedge and label the intersection as B.
Set your compass to the distance from C to B. Position it on B and mark an arc on EB. Designate this intersection point as A.
Thus, AB will equal both CB and BD.
In this scenario, we start with the following expression:

The first step involves evaluating the quadratic component.
Next, we have:

The following step is to subtract the two resultant values:

It is clear that the outcome of this operation is a negative value.
Answer:
The expression results in:

We recognize that two angles, ∠UVW and ∠XYZ, are complementary, which means their sum is 90°.
Their measures are given as:
∠UVW = x - 10
∠XYZ = 4x - 10
Adding these, we have:
(x - 10) + (4x - 10) = 90
Simplifying:
5x - 20 = 90
Adding 20 to both sides:
5x = 110
Dividing by 5:
x = 22
Substituting back:
∠UVW = 22 - 10 = 12°
∠XYZ = 4(22) - 10 = 78°
Therefore, the values are:
x = 22°
∠UVW = 12°
∠XYZ = 78°
Answer:
50
Step-by-step clarification:
The equation that represents the total cost is in the format of a linear equation y = mx + c
Here, m signifies the slope of the line
c indicates the y-intercept, showing where the line intersects the y-axis
When the equation y = 150x + 50 is plotted, it will form a linear graph where the y-intercept corresponds to 50, as observed in the standard form of a linear equation.
Answer:
1. What are the amplitude and period of the sine function that indicates the positioning of the band members as they start performing?
Answer: The amplitude is 80 ft and the period is 60 ft.
2. Edna, seated in the stands, faces Darla and notices that the sine curve starts rising from the left edge of the field. What is the equation for the sine function representing the arrangement of band members at the beginning of their performance?
Answer: y = 80cos(x*π/30)+80
3. When the band starts playing, the members move away from the edges, and the sine curve changes to start decreasing at the far left. Darla remains in her position. Now the sine curve is half as tall as it originally was. What is the equation for the updated sine curve depicting the band members' positions?
Answer: y = 40cos(x*π/30)+80
4. Finally, the entire band shifts closer to the edge of the football field, causing the sine curve to now position itself in the lower half of the field from Edna’s perspective. What is the equation for this sine curve reflecting the band members' positions after these adjustments?
Answer: y = 40cos(x*π/30)+40
Step-by-step explanation: