Width =3.5, length = 4
Let's make a few equations to summarize what we know.:
a = 14
l = 2w -3
a = wl
Now let's substitute the expression for l into the expression for area and solve.
a = w(2w - 3)
a = 2w^2 - 3w
14 = 2w^2 - 3w
0 = 2w^2 - 3w - 14
which factors into
(2w - 7) (w + 2)
Solving for w
2w - 7 = 0
2w = 7
w = 3.5
w+2 = 0
w = -2
So w can be either 3.5 or -2. Since a negative width doesn't make sense, w = 3.5
Let's verify
w = 3.5
l = 2w - 3
l = 2*3.5 - 3
l = 7 - 3
l = 4
a = wl
a = 3.5 * 4
a = 14
So the width is 3.5 and the length is 4.
<span>Answer:
K = (1/2) mv² + (1/2) Iω², where m is the ball mass, I is the ball's moment of inertia (2/5)mr², and ω is the angular velocity of the ball. Because the ball rolls without slipping, it is easy to see that v=ωr, or r=v/ω. Then,
K = (1/2)mv² + (1/2)(2/5)mr²ω²
= (1/2)mv² + (1/5)mv²
= (7/10)mv²
Setting potential at the top equal to kinetic at the bottom,
mgh=(7/10)mv²
v=âš{(10/7)(gh)}
= [(10/7)(9.8)(0.51)]^(1/2) = 2.672m/s</span>
Answer:
3x + 3
Step-by-step explanation:
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.