Which sequence of transformations will map figure K onto figure K′? Two congruent kites, figure K and figure K prime, are drawn
on a coordinate grid. Figure K has vertices at 4, 3, at 6, 5, at 4, 8, and at 2, 5. Figure K prime has vertices at 4, negative 8, at 6, negative 5, at 4, negative 3, and at 2, negative 5 Reflection across x = 4, 180° rotation about the origin, and a translation of (x + 8, y) Reflection across x = 4, 180° rotation about the origin, and a translation of (x − 8, y) Reflection across y = 4, 180° rotation about the origin, and a translation of (x + 8, y) Reflection across y = 4, 180° rotation about the origin, and a translation of (x − 8, y)
2 answers:
Answer:
The sequence of transformations will map figure K onto figure K′
is the first sequence <u>option (1)</u>
======================================================
Step-by-step explanation:
See the attached figure
as shown in the figure the K' is the image of K by reflection over x-axis
<u>But </u> We need to know which sequence of transformations will give the same result.
So, we will test the options by any point from K and its image from K'
i.e: we will test the options using the points (6,5) , (6,-5)
(6,5) ⇒ (6,-5)
<u>option (1):</u>
Reflection across x = 4, 180° rotation about the origin, and a translation of (x + 8, y)
(6,5) ⇒ (2,5) ⇒(-2,-5) ⇒ (6,-5)
<u>option (2):</u>
Reflection across x = 4, 180° rotation about the origin, and a translation of (x − 8, y)
(6,5) ⇒ (2,5) ⇒(-2,-5) ⇒ (-10,-5)
<u>option (3):</u>
Reflection across y = 4, 180° rotation about the origin, and a translation of (x + 8, y)
(6,5) ⇒ (6,3) ⇒ (-6,-3) ⇒ (2,-3)
<u>option (4):</u>
Reflection across y = 4, 180° rotation about the origin, and a translation of (x − 8, y)
(6,5) ⇒ (6,3) ⇒ (-6,-3) ⇒ (-14,-3)
As shown: The sequence of transformations will map figure K onto figure K′
is the first sequence <u>option (1)</u>
You might be interested in
we have
where
c---------> is the total cost of going to the carnival
t---------> is the number of tickets purchased
we know that
<u>the domain of the function is the interva</u>l---------------> [0,∞)
Because the number of tickets can not be negative
<u>The range of the function is the interval</u>-------------> [10,∞)
Because the total cost can not be negative
using a graph tool
see the attached figure
The answer in the attached figure
Answer:
If thrown up with the same speed, the ball will go highest in Mars, and also it would take the ball longest to reach the maximum and as well to return to the ground.
Step-by-step explanation:
Keep in mind that the gravity on Mars; surface is less (about just 38%) of the acceleration of gravity on Earth's surface. Then when we use the kinematic formulas:
the acceleration (which by the way is a negative number since acts opposite the initial velocity and displacement when we throw an object up on either planet.
Therefore, throwing the ball straight up makes the time for when the object stops going up and starts coming down (at the maximum height the object gets) the following:
When we use this to replace the 't" in the displacement formula, we et:
This tells us that the smaller the value of "g", the highest the ball will go (g is in the denominator so a small value makes the quotient larger)
And we can also answer the question about time, since given the same initial velocity , the smaller the value of "g", the larger the value for the time to reach the maximum, and similarly to reach the ground when coming back down, since the acceleration is smaller (will take longer in Mars to cover the same distance)