Given: ∠ DEF
To construct: ∠TSZ ≅ ∠DEF
Construction: Consider the attachment
Step-01: Draw a line XY and choose a point S on it as a vertex of the required angle. Further marks point T such that DE = ST
Step-02: Take an arc AB from point E in ∠DEF of any length and draw at point S which cuts at point P on XY line.
Step-03: Take another arc of length AB from point B in ∠DEF and draw from point P which cuts to the previous arc at Q.
Step-04: Now, join the point SQ and extend up to Z such that EF = SZ
Hence, ∠ TSZ will be the required congruent constructed angle to∠DEF
Answer:
Step-by-step explanation:
So we have the equation:
And we want to solve for g.
First, isolate g. To do so, subtract vt from both sides:
Multiply both sides by 2:
Now, divide both sides by t^2:
Expand:
Simplify the second term:
And we're done!
Answer:
0ft and 60ft
Step-by-step explanation:
Given
The attached function
Required
Determine the valid values of the domain of the function
To do this, we simply consider the starting point and the end point of the trajectory on the x-axis (i.e. the horizontal distance).
From the attached graph, the horizontal distance starts from 0 and ends at 180.
This implies that the domain is: 
From the options, the values that fall in this bracket are 0ft and 60ft
D. (5,-21) would be your answer
Keep in mind x >0 since x is a measurement of length.
