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Ugo [173]
2 years ago
11

Parallelogram FGHJ was dilated and translated to form

Mathematics
2 answers:
Veronika [31]2 years ago
8 0

Answer:

4 on edge

Step-by-step explanation:

Mrrafil [7]2 years ago
3 0

Answer:  The required scale factor of the dilation is 4.  

Step-by-step explanation:  Given that the parallelogram FGHJ was dilated and translated to form similar parallelogram F'G'H'J'.

We are to find the scale factor of the dilation.

From the graph, we note that

JH = 2 units   and   J'H' = 8 units.

We know that

\textup{Scale factor of dilation}=\dfrac{\textup{length of a side of the dilated figure}}{\textup{length of the correponding side of the original figure}}.

Therefore, the scale factor of the given dilation is

S=\dfrac{J'H'}{JH}\\\\\\\Rightarrow S=\dfrac{8}{2}\\\\\Rightarrow S=4.

Thus, the required scale factor of the dilation is 4.  

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Mikhail believes he can use the multiplication expression 1/2x6 to find the total number of
Svetlanka [38]

Answer:

I think the second option is the most appropriate option.

Step-by-step explanation:

Mikhail believes he can use the multiplication expression \frac{1}{2} \times 6 to find the total number of  pound servings of rice he can make  from 6 pounds of rice.

Now, we have to select the statement from the four options that best explains why Mikhail is correct or incorrect.

I think the second option is the most appropriate option.

So, Mikhail is correct in using a multiplication expression because 1/2 is the number of servings per pound. (Answer)

7 0
2 years ago
The number of hours a group of contestants spends preparing for a quiz show is listed below. What is the frequency table that re
harkovskaia [24]
I'll just show you how to make a frequency table using the above data. 
We will group the data into class intervals and determine the frequency of the group.

<span>8 12 25 32 45 50 62 73 80 99 4 18 9 39 36 67 33
</span>
smallest data value = 4
highest data value = 99
difference = 99 - 4 = 95
number of data = 17

Let us assign a class interval of 20.

Class Interval     Tally                              Frequency
0-20                    8, 12, 4, 18, 9,                     5
21-40                25, 32, 39, 36, 33                 5 
41-60                45, 50, 67                              3
61-80                62, 73, 80                              3
81-100              99                                           1

That is how a frequency table look like. Usually, under the Tally column, tick marks are written instead of the numbers but for easier monitoring, I used the numbers in the data set.
8 0
2 years ago
Mr. Phillips is mixing paint for his art class. How many 6-ounce bottles can he fill with the quantities of paint shown at the r
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Is there any images?
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2 years ago
Which quadratic equation is equivalent to (x2 – 1)2 – 11(x2 – 1) + 24 = 0? u2 – 11u + 24 = 0 where u = (x2 – 1) (u2)2 – 11(u2) +
Vinvika [58]
<span><span>u2</span> – 11u + 24 = 0 where u = (x2 – 1) the first awnser, a.</span>
4 0
2 years ago
Read 2 more answers
Roberto wants to display his 18 sports cards in an album. Some pages hold 2 cards and others hold 3 cards. How many different wa
WITCHER [35]
<span>65 = number of different arrangements of 2 and 3 card pages such that the total number of card slots equals 18. 416,154,290,872,320,000 = number of different ways of arranging 18 cards on the above 65 different arrangements of page sizes. ===== This is a rather badly worded question in that some assumptions aren't mentioned. The assumptions being: 1. The card's are not interchangeable. So number of possible permutations of the 18 cards is 18!. 2. That all of the pages must be filled. Since the least common multiple of 2 and 3 is 6, that means that 2 pages of 3 cards can only be interchanged with 3 pages of 2 cards. So with that said, we have the following configurations. 6x3 card pages. Only 1 possible configuration. 4x3 cards and 3x2 cards. These pages can be arranged in 7!/4!3! = 35 different ways. 2x3 cards and 6x2 cards. These pages can be arranged in 8!/2!6! = 28 ways 9x2 card pages. These can only be arranged in 1 way. So the total number of possible pages and the orders in which that they can be arranged is 1+35+28+1 = 65 possible combinations. Now for each of those 65 possible ways of placing 2 and 3 card pages such that the total number of card spaces is 18 has to be multiplied by the number of possible ways to arrange 18 cards which is 18! = 6402373705728000. So the total amount of arranging those cards is 6402373705728000 * 65 = 416,154,290,872,320,000</span>
6 0
2 years ago
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