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Anarel [89]
3 years ago
13

A quadrilateral with vertices (-3,2),(-5,-4), (4,6), and (7,0) is dilated by a scale factor of 0.5.

Mathematics
1 answer:
Ksenya-84 [330]3 years ago
6 0
Original vertices:
1 (-3, 2)
2 (-5,-4)
3 ( 4, 6)
4 ( 7, 0)

dilated by a scale factor of 0.50

1) -3*0.50 = -1.5 ; 2 * 0.50 = 1 ⇒ (-1.5,1)
2) -5*0.50 = -2.5 ; -4*0.50 = -2 ⇒(-2.5,-2)
3) 4*0.50 = 2; 6*0.50 = 3 ⇒ (2,3)
4) 7*0.50 = 3.5; 0*0.50 = 0 ⇒ (3.5,0)

B.)
 -1.5    -2.5    2    3.5 
<span>     1       -2    3       0 </span>

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If XY × 8 = YYY, where Xand Y are digits of the numbers, then what is the value of Y?
kakasveta [241]
Given that
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and Y is one of the digits from 1 to 9
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YYY = 100Y + 10Y + Y = Y(100+10+1) = 111Y
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substitute with X = 1 to 9
X = 1 ⇒⇒⇒ Y = 0.77 ⇒⇒ unacceptable

X = 2 ⇒⇒⇒ Y = 1.55 ⇒⇒ unacceptable
X = 3 ⇒⇒⇒ Y = 2.33 ⇒⇒ unacceptable
X = 4 ⇒⇒⇒ Y = 3.11 ⇒⇒ unacceptable
X = 5 ⇒⇒⇒ Y = 3.88 ⇒⇒ unacceptable
X = 6 ⇒⇒⇒ Y = 4.66 ⇒⇒ unacceptable
X = 7 ⇒⇒⇒ Y = 5.44 ⇒⇒ unacceptable
X = 8 ⇒⇒⇒ Y = 6.21 ⇒⇒ unacceptable

X = 9 ⇒⇒⇒ Y = 6.99 ⇒⇒ unacceptable

So, The is no value of  Y to achieve ⇒⇒ XY * 8 = YYY

================================================

I think the problem is as following:

Given that XY8 = YYY ⇒⇒⇒ Where X and Y are digits
So, X is equal to one of the digits from 1 to 9
and Y is one of the digits from 1 to 9
This can be solved as following
YYY = 100Y + 10Y + Y = Y(100+10+1) = 111Y
XY8 = 100X + 10Y + 8
∴ 100X + 10Y + 8 = 111Y
∴ 100x + 8 = 101Y
∴ Y = (100X + 8)/101
substitute with X = 1 to 9
X = 1 ⇒⇒⇒ Y = 1.07 ⇒⇒ unacceptable

X = 2 ⇒⇒⇒ Y = 2.06 ⇒⇒ unacceptable
X = 3 ⇒⇒⇒ Y = 3.05 ⇒⇒ unacceptable
X = 4 ⇒⇒⇒ Y = 4.04 ⇒⇒ unacceptable
X = 5 ⇒⇒⇒ Y = 5.03 ⇒⇒ unacceptable
X = 6 ⇒⇒⇒ Y = 6.02 ⇒⇒ unacceptable
X = 7 ⇒⇒⇒ Y = 7.01 ⇒⇒ unacceptable
X = 8 ⇒⇒⇒ Y = 8    ⇒⇒⇒ integer ⇒⇒ the correct answer

X = 9 ⇒⇒⇒ Y =8.99 ⇒⇒ unacceptable

So, The value of Y = 8
8 0
3 years ago
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