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Nuetrik [128]
3 years ago
14

Segment A prime B prime has endpoints located at A' (0, 0) and B' (2, 0). It was dilated at a scale factor of one half from cent

er (2, 0). Which statement describes the pre-image?
has endpoints located at A' (0, 0) and B' (2, 0). It was dilated at a scale factor of one half from center (2, 0). Which statement describes the pre-image?
segment AB is located at A (−2, 0) and B (2, 0) and is half the length of segment A prime B prime.
segment AB is located at A (−2, 0) and B (2, 0) and is twice the length of segment A prime B prime.
segment AB is located at A (−1, 0) and B (1, 0) and is half the length of segment A prime B prime.
segment AB is located at A (−1, 0) and B (1, 0) and is twice the length of segment A prime B prime.
Mathematics
2 answers:
Ira Lisetskai [31]3 years ago
6 0

Answer:Aubrey is correct i took the test and got it right

Step-by-step explanation:

yarga [219]3 years ago
4 0

Answer:

i believe the answer is segment AB is located at A (-2, 0) and B (2,0) and is twice the length of segment A prime B prime

Step-by-step explanation:

^^^

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It’s 5/4. Your welcome
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3 years ago
The ratio of men to women in a movie theater is 3:5. As two more men and one more woman walk in, the ratio of men to women becom
34kurt

Answer:

C. 20

Step-by-step explanation:

Let's say M is the original number of men and W is the original number of women.

M / W = 3 / 5

(M+2) / (W+1) = 2 / 3

Let's cross multiply both equations:

5M = 3W

3(M+2) = 2(W+1)

Let's simplify the second equation:

3M + 6 = 2W + 2

3M + 4 = 2W

From the first equation:

M = 3/5 W

Substitute:

3 (3/5 W) + 4 = 2W

9/5 W + 4 = 2W

4 = 1/5 W

W = 20

There were originally 20 women.

Let's check our answer.  That would mean that M = 3/5 W = 12.

After 2 men walk in and 1 woman, W = 21 and M = 14, so 14/21 = 2/3.  Looks like the answer is correct!

Answer C.

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4 years ago
There are 30 boys in 6th grade. The number of girls in 6th grade is 40. Abigail says that means the ratio of the number of boys
Kipish [7]
Yes She is Correct .
3 0
3 years ago
90 points!!
Svetllana [295]

Answer:

1) 2w + 2(3w) \leq 112

2) 850w > 300w + 7500

3) The greatest age that Sue could be is 7.

4) The smaller of the two integers is 92.

5) Don needs to earn at least 244 points in the fourth game.

Step-by-step explanation:

1) An rectangle has 2 dimensions: width(w) and length(l)

The perimeter P is:

P = 2w + 2l

The problem states that the length of a rectangle is three times its width. So l = 3w and:

P = 2w + 2(3w)

The perimeter of the rectangle is at most 112 cm. It means that the perimeter can be 112, so the equal sign enters the inequality. So

2w + 2(3w) \leq 112

2)

The problem states that he earns $850 per week in sales. His earnings is modeled by the following equation:

E = 850*w, in which w is the number of weeks.

The problem also states that he spent $7500 to obtain his merchandise, and it costs him $300 per week for general expenses. So his expenses can be modeled by the following equation

C = 300*w + 7500, in which w is also the number of weeks.

He will make a profit when his earnings are bigger than his expenses, so: When they are equal, there is no profit, so the equal sign does not enter the inequality.

E > C

850w > 300w + 7500

3)

I am going to call Jenny's age x and Sue's age y.

The problem states that Jenny is eight years older than twice her cousin Sue’s age. So

x = 8 + 2y.

The sum of their ages is less than 32, so:

x + y < 32

8 + 2y + y < 32

3y < 24

y < \frac{24}{3}

y < 8

Sue's age has to be less than 8, so the greatest age that Sue could be is 7.

4)

The sum of two consecutive integers is at least 185.

There are two integers with sum of 185, so:

x + y = 185

They are consecutive so:

x = y + 1

Replacing in the sum equation:

y + 1 + y = 185

2y = 184

y = \frac{184}{2}

y = 92

The smaller of the two integers is 92.

5)

The average is the sum of all the scores divided by the number of games. So:

225 = \frac{192 + 214 + 250 + x}{4}

656 + x = 900

x = 900 - 656

x = 244

Don needs to earn at least 244 points in the fourth game.

8 0
4 years ago
A student wanted to find the sum of all the even numbers from 1 to 100. He said:The sum of all the even numbers from 1 to 100 is
hram777 [196]
There would only be 50 odd numbers between 1 and 100 so therefor there would be 50 even so that would only be 50 odd numbers 
4 0
3 years ago
Read 2 more answers
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