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Arte-miy333 [17]
2 years ago
5

Which of these sequences of transformations would not return a shape to its original position?

Mathematics
1 answer:
oksano4ka [1.4K]2 years ago
7 0

Answer:

4.  Rotate 120∘ counterclockwise around center C, then rotate 220∘ counterclockwise around C again.

Step-by-step explanation:

Option 1, 2 and 3 will return a shape back to its original because:

1. 3 units up and 3 units down are direct opposite and will cancel out one another. Hence, the shape will return to its initial position.

2. Reflect over line p, and over p again.

This actions are also direct opposite and also have the same effect as (1)

3. Translate 1 unit to the right and 3 units to the right.

This equates to 4 units to the right and it will cancel out the translation of 4 units to the left.

4. Rotate 120 counter clockwise and another 220 counterclockwise will not make a shape go back to its original position.

Hence:

4 answers the question

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tensa zangetsu [6.8K]
Well 13 /3 = 4.3333, that sure the cost for one book and the rate. so if he bought 12 books at the same rate, then its 12*4.33333. which equals 52 dollars
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2 years ago
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List all of the possible outcomes using the table below.
NemiM [27]

Answer:

The four possible outcomes are:

  • skirt 1, white
  • skirt 1, black
  • skirt 2, white
  • skirt 2, black

Step-by-step explanation:

Skirt 1 can be white or black (two outcomes). Skirt 2 can be white or black (two outcomes). Therefore, there are four possible outcomes.

6 0
2 years ago
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PLEASE HELP!!!!!!!!!!!!! DUE SOON
Sergio039 [100]

\bf ~~~~~~\textit{initial velocity} \\\\ \begin{array}{llll} ~~~~~~\textit{in feet} \\\\ h(t) = -16t^2+v_ot+h_o \end{array} \quad \begin{cases} v_o=\stackrel{}{\textit{initial velocity of the object}}\\\\ h_o=\stackrel{}{\textit{initial height of the object}}\\\\ h=\stackrel{}{\textit{height of the object at "t" seconds}} \end{cases} \\\\[-0.35em] ~\dotfill\\\\ h=-16t^2+\stackrel{\stackrel{v_o}{\downarrow }}{65}t


now, take a look at the picture below, so for 2) and 3) is the vertex of this quadratic equation, 2) is the y-coordinate and 3) the x-coordinate.


\bf \textit{vertex of a vertical parabola, using coefficients} \\\\ h=\stackrel{\stackrel{a}{\downarrow }}{-16}t^2\stackrel{\stackrel{b}{\downarrow }}{+65}t\stackrel{\stackrel{c}{\downarrow }}{+0} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right) \\\\\\ \left( -\cfrac{65}{2(-16)}~~,~~0-\cfrac{65^2}{4(-16)} \right) \implies \left( \cfrac{65}{32}~,~0- \cfrac{4225}{-64}\right)


\bf \left( \cfrac{65}{32}~,~0+ \cfrac{4225}{64}\right)\implies \left( \stackrel{seconds}{2\frac{1}{32}}~~,~~ \stackrel{feet~hight}{66\frac{1}{64}}\right)

6 0
2 years ago
1. What is the theoretical probability that the family has two dogs or two cats?
gogolik [260]

let dogs be heads. Let cats be tails. A coin has two sides, in which you are flipping two of them. Note that there can be the possible outcomes  

h-h, h-t, t-h, t-t.  

How this affects the possibility of two dogs & two cats. Note that there are 1/2 a chance of getting those two (with the others being one of each), which means that out of 4 chances, 2 are allowed.  

2/4 = 1/2  

50% is your answer

Heads represents cats and tails represents dogs. There is two coins because we are checking the probability of two pets. You have to do the experiment to get your set of data, once you get your set of data, you will be able to divide it into the probability for cats or dogs. To change the simulation to generate data for 3 pets, simply add a new coin and category for the new pet.

Hope this helps you out!

4 0
3 years ago
What is the value of p in the equation 13p+42=-57+2p
ira [324]

Answer:

p=-9

Step-by-step explanation:

13p+42=-57+2p

Subtract 2p from each side

13p-2p+42=-57+2p-2p

11p +42 = -57

Subtract 42 from each side

11p+42-42=-57-42

11p = -99

Divide each side by 11

11p/11 = -99/11

p = -9

6 0
3 years ago
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