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Basics of Transformations
Students were asked to create true statements about transformations. Circle the names of the students who correctly completed the task. Then, unscramble the underlined letters of the circled names to answer riddle of the bottom.
Answers
Answer:
The 6 names to circle are:
LavernaAlfonsoAutumnIgnacioKathyrnLawrenceThe underlined letters from those circled names are:
RNONTUGRNWRThe letters form the phrase [tex]\underline{\text{WR}}\ \underline{\text{ON}} \ \underline{\text{G}} \ \ \ \ \ \ \underline{\text{TU}} \ \underline{\text{RN}}[/tex] or "wrong turn". The spacing is done like that in the first version to show the various subgroups of letters.
Unfortunately the mentioned riddle at the bottom is cut off, so I don't know what the riddle is.
======================================================
Explanation:
Laverna is correct because a translation is applied. "Translation" in geometry settings means "shifting up/down/left/right". In this case, triangle JKL is shifted 5 units right and 3 units up.Alfonso is correct. If a figure is labeled ABCD going in clockwise orientation, then a translation will preserve said orientation. The orientation only flips when a reflection occurs.Autumn is correct. Why? Because reflections are isometries, meaning they preserve distances and lengths and angles. In other words, the two triangles are twin clones of each other.Willa is not correct. The prime notations go on the image and NOT the preimage. In other words, the prime notations go for the "after" and not the "before". Example: triangle JKL in Laverna's diagram is the preimage, while J'K'L' is the image. JKL is "before", and J'K'L' is "after".Ignacio is correct. The figure enlarges or shrinks (depending on the scale factor if its larger than 1, or smaller than 1). The orientation stays the same. Refer to Alfonso's scenario where I mentioned the orientation flipping only when a reflection happens.Napoleon is incorrect. A reflection ALWAYS changes the orientation. Consider a triangle ABC where the motions from A to B to C goes clockwise. A reflection will flip the orientation to make A to B to C go counterclockwise. Example: Reflections over non-horizontal lines swap the positions of left vs right, which is another way to see why the orientation swaps as well.Kathyrn is correct assuming no reflection operations are done.Titus is not correct. A rotation is not the same as a reflection. Though I should point out that two reflections simplify to a rotation. The mirror lines must intersect in some way. Parallel mirror lines will have two reflections lead to a translation.Lawrence is correct. The order P,Q,R is clockwise, and so is the order P', Q', R'. A rotation preserves orientation. In this case, a 180 degree rotation has been done. The rule for that is any (x,y) point turns into (-x,-y).Here's a summary table:
[tex]\begin{array}{|c|c|} \cline{1-2}\text{Student} & \text{Are They Correct?}\\\cline{1-2}\text{Laverna} & \textbf{Yes}\\\cline{1-2}\text{Alfonso} & \textbf{Yes}\\\cline{1-2}\text{Autumn} & \textbf{Yes}\\\cline{1-2}\text{Willa} & \text{No}\\\cline{1-2}\text{Ignacio} & \textbf{Yes}\\\cline{1-2}\text{Napoleon} & \text{No}\\\cline{1-2}\text{Kathyrn} & \textbf{Yes}\\\cline{1-2}\text{Titus} & \text{No}\\\cline{1-2}\text{Lawrence} & \textbf{Yes}\\\cline{1-2}\end{array}[/tex]
Put another way, here are the people to circle (i.e. the people who made true statements):
LavernaAlfonsoAutumnIgnacioKathyrnLawrenceIn the order presented above, the letters underlined are:
RNONTUGRNWRThrough a bit of trial and error, the letters make the phrase of:
[tex]\underline{\text{WR}}\ \underline{\text{ON}} \ \underline{\text{G}} \ \ \ \ \ \ \underline{\text{TU}} \ \underline{\text{RN}}[/tex]
I'm not sure what to do with the extra "RN". It might be a typo, or it might be the case your teacher wants you to ignore repeats like this.
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Hi Student!
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Answers
Answer:
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Find the probability that a randomly
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Answer:
[tex]31.8\%[/tex]
Step-by-step explanation:
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Answer:
a. [tex]x^{-2}[/tex]
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Step-by-step explanation:
A blimp, suspended in the air at a height of 600 feet, lies directly over a line from a sports stadium to a planetarium. If an angle of depression from the blimp to the stadium is 37 degrees and from the blimp to the planetarium is 29 degrees, find the distance between the sports stadium and the planetarium.
Answers
Step-by-step explanation:
so, this sounds like the blimp is located between the stadium and the planetarium.
we have a triangle :
the ground distance between the stadium and the planetarium is the baseline.
and the 2 lines of sight from the blimp on one side to the stadium and on the other side to the planetarium are the 2 legs.
we know the height of this triangle is 600 ft.
the angle of depression down to the stadium is 37°. which makes the inner triangle angle at the ground point at the stadium also 37°.
and the angle of depression down to the planetarium is 29°. which makes the inner triangle angle at the ground point at the planetarium also 29°.
and because the sum of all angles in a triangle is always 180°, we know the angle at the blimp is
180 - 37 - 29 = 114°
in order to solve this triangle, we need to split it into 2 right-angled triangles by using the height of the main triangle as delimiter.
we get a stadium side and a planetarium side triangle.
the baselines (Hypotenuses) of the 2 triangles are the corresponding lines of sight from the blimp.
the height of the large triangle is also a height and a leg in each small triangle.
and the stadium side part of the large baseline (between ground point and intersection with the height) is the second leg for the stadium side triangle.
and correspondingly, the planetarium side part of the large baseline (between ground point and intersection with the height) is the second leg for the planetarium side triangle.
the inner blimp angle of the stadium side triangle is
180 - 37 - 90 = 53°
and the inner blimp angle of the planetarium side triangle is
180 - 29 - 90 = 61°
now we can use the law of sine to get the lengths of the 2 parts of the baseline of the large triangle.
and when we add these 2 numbers we get the distance between stadium and planetarium.
law of sine is
a/sin(A) = b/sin(B) = c/sin(C)
with the sides being opposite of the associated angles.
for the stadium side triangle we get
part1/sin(53) = 600/sin(37)
part1 = 600×sin(53)/sin(37) = 796.226893... ft
for the planetarium side triangle we get
part2/sin(61) = 600/sin(29)
part2 = 600×sin(61)/sin(29) = 1,082.428653... ft
the distance between the stadium and the planetarium is
part1 + part2 = 1,878.655546... ft
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This above implies that:
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The value of 2.4494897… is equivalent to √6.
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Answer:
is it 1:4 or am I missing something
Step-by-step explanation:
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Answer:
Simplifying
-9 + 4 = 40
Combine like terms: -9 + 4 = -5
-5 = 40
Solving
-5 = 40
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Your history book cost $43 dollars in 1999. The earlier edition sold for $37 in 1992. Comparing the converted costs, which book is more expensive? (Assume the 1999 CPI is 166.3 and the 1992 CPI is 140.3)
Answers
When the converted costs are compared, the earlier edition of the history book is more expensive.
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Converted price of the earlier edition = (CPI in 1992 / CPI in 1999) x later edition of the book
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