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Blizzard [7]
3 years ago
8

30 POINTS,NEED HELP ASAP !!!

Mathematics
2 answers:
zaharov [31]3 years ago
7 0

Answer: OPTION A.

Step-by-step explanation:

You can observe that in the figure CDEF the vertices are:

C(-2,-1),\ D(-2,0),\ E(2,2)\ and\ F(2,1)

And in the figure C'D'E'F'  the vertices are:

C'(-8,-4),\ D'(-8,0),\ E'(8,8)\ and\ F'(8,4)

For this case, you can divide any coordinate of any vertex of the figure C'D'E'F' by any coordinate of any vertex of the figure CDEF:

For C'(-8,-4) and C(-2,-1):

\frac{-8}{-2}=4\\\\\frac{-4}{-1}=4

Let's choose another vertex. For E'(8,8) and E(2,2):

\frac{8}{2}=4\\\\\frac{8}{2}=4

You can observe that the coordinates of C' are obtained by multiplying each coordinate of C by 4 and the the coordinates of E' are also obtained by multiplying each coordinate of E by 4.

Therefore, the rule that yields the dilation of the figure CDEF centered at the origin is:

(x, y)→(4x, 4y)

SashulF [63]3 years ago
5 0

Answer:

A. (x, y) => (4x, 4y) this will help you out ;-)

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Answer:

Step-by-step explanation:

Based on the question we are given the percentages of each of the types of candies in the bag except for brown. Since the sum of all the percentages equals 75% and brown is the remaining percent then we can calculate that brown is (100-75 = 25%) 25% of the bag. Now we can show the probabilities of getting a certain type of candy by placing the percentages over the total percentage (100%).

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  • Yellow or Blue: \frac{20}{100} +\frac{20}{100} = \frac{40}{100}  ....add the numerators
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  • Stiped:  \frac{25}{100} .... there are 0 striped candies.

Assuming the <u><em>ratios/percentages</em></u> of the candies stay the same having an infinite amount of candy will not affect the probabilities. That being said in order to calculate consecutive probability of getting 3 of a certain type in a row we have to multiply the probabilities together. This is calculated by multiplying the numerators with numerators and denominators with denominators.

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  • the 1st and 3rd are red while the middle is any. We multiply 15% * (total of all minus red which is 85%) * 15% like so.

\frac{15*85*15}{100*100*100} = \frac{19,125}{1,000,000} = \frac{1.9125}{100}

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\frac{80*80*80}{100*100*100} = \frac{512,000}{1,000,000} = \frac{51.2}{100}

  • At least 1 green: multiply the percent of green by 100% twice, since the other two can by any

\frac{20*100*100}{100*100*100} = \frac{200,000}{1,000,000} = \frac{20}{100}

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