B and D because it doesnt matter how you write the equation you still get the same answer,
don't forget to rate my answer 5 stars and thank me
Answer:
- square: 9 square units
- triangle: 24 square units
Step-by-step explanation:
Using a suitable formula the area of a polygon can be computed from the coordinates of its vertices. You want the areas of the given square and triangle.
<h3>Square</h3>
The spreadsheet in the first attachment uses a formula for the area based on the given vertices. It computes half the absolute value of the sum of products of the x-coordinate and the difference of y-coordinates of the next and previous points going around the figure.
For this figure, going to that trouble isn't needed, as a graph quickly reveals the figure to be a 3×3 square.
The area of the square is 9 square units.
<h3>Triangle</h3>
The same formula can be applied to the coordinates of the vertices of a triangle. The spreadsheet in the second attachment calculates the area of the 8×6 triangle.
The area of the triangle is 24 square units.
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<em>Additional comment</em>
We have called the triangle an "8×6 triangle." The intention here is to note that it has a base of 8 units and a height of 6 units. Its area is half that of a rectangle with the same dimensions. These dimensions are readily observed in the graph of the vertices.
Don't be worried friend :)
-(y + 2) + 8 = 3
=> -(y + 2) = -5
=> y + 2 = 5
=> y = 3
1/9 times 2/2=2/18
2/3 times 6/6=12/18
2/18+12/18+5/18=(2+12+5)/18=19/18=1 and 1/8
A is answer
If y = cos(kt), then its first two derivatives are
y' = -k sin(kt)
y'' = -k² cos(kt)
Substituting y and y'' into 49y'' = -16y gives
-49k² cos(kt) = -15 cos(kt)
⇒ 49k² = 15
⇒ k² = 15/49
⇒ k = ±√15/7
Note that both values of k give the same solution y = cos(√15/7 t) since cosine is even.