For each problem, find the missing value using the Pythagorean Theorem
1 answer:
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Answer:
We would have
where " l " is length, " w" is width and "h" is height.
Step-by-step explanation:
Step 1
Remember that
Surface area for a box with no top =
where " l " is length, " w" is width and "h" is height.
Step 2.
Remember as well that
Volume of the box =
Step 3
We can now use lagrange multipliers. Lets say,
and
By the lagrange multipliers method we know that
Step 4
Remember that
and
So basically you will have the system of equations
Now, remember that you can multiply the first eqation, by "l" the second equation by "w" and the third one by "h" and you would get
Then you would get
You can get rid of from these equations and you would get
And from those equations you would get
Now remember the original equation
If we plug in what we just got, we would have
Answer:
The image triangle ΔD'L'W' after the dilation by a scale factor of 1/2 will have the vertices:
- D' (0, 1)
- L' (1/2, 5/2) or L'(0.5, 2.5)
- W' (5/2, 2) or W'(2.5, 2)
`The figure of the dilated image is attached below.
Step-by-step explanation:
Given the triangle ΔDLW withe the vertices
- D(0, 2)
- L(1, 5)
- W(5, 4)
As the scale factor 1/2 < 0, thus the image triangle will be smaller than the original triangle.
The rule of dilation is given by
Dilation: (x, y) → (1/2x, 1/2y), centered at (0, 0)
D(0, 2) → D' (1/2x, 1/2y) → D' (1/2(0), 1/2(2)) = D' (0, 1)
L(1, 5) → L'(1/2x, 1/2y) → L'(1/2(1), 1/2(5)) = L' (1/2, 5/2)
W(5, 4) → W'(1/2x, 1/2y) → W'(1/2(5), 1/2(4)) = W' (5/2, 2)
Therefore, the image triangle ΔD'L'W' after the dilation by a scale factor of 1/2 will have the vertices:
- D' (0, 1)
- L' (1/2, 5/2) or L'(0.5, 2.5)
- W' (5/2, 2) or W'(2.5, 2)
`The figure of the dilated image is attached below.
