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Luden [163]
2 years ago
10

AB is dilated by a scale factor of 3 to form A'B' . Point O, which lies on AB , is the center of dilation. The slope of AB is 3.

The slope of A'B' is . through point O.
Mathematics
2 answers:
Anna [14]2 years ago
7 0
Given that AB has been dilated by scale factor 3 to form A'B', and by laws of dilation, the image is congruent to the pre image. This implies that the image will not change in any way whatsoever apart from the size. Hence the slope will remain the same. That means the slope of A'B' will be 3
Triss [41]2 years ago
7 0

Answer:

The slope of A'B'=3

Step-by-step explanation:

We are given that AB is dilated by scale factor of 3 to form A'B'.

A point O lies on  the line AB which is the center of dilation.

The slope of AB=3

We have to find the slop A'B' .

We know that

Dilation is that transformation in which shape of figure does not change but the size of figure changes. Pre-image and image obtained after dilation are similar.

Therefore, slope of A'B' does not change.

The slope of AB =Slope of A'B'=3

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Step-by-step explanation:

Simply multiply the first question by 12 to get 2x+8y=96

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Suppose that the IQs of university​ A's students can be described by a normal model with mean 140 and standard deviation 8 poi
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Answer:

0.0266, 0.9997,0.7856

Step-by-step explanation:

Given that the IQs of university​ A's students can be described by a normal model with mean 140 and standard deviation 8 points. Also suppose that IQs of students from university B can be described by a normal model with mean 120 and standard deviation 11. Let x be the score by A students and Y the score of B.

A)P(X>135) = \\P(Z>0.625)\\=0.0266

B) Since X and Y are independent we have

X-Y is Normal with mean = 140-120 =20 and Var (x-y)=Var(x)+Var(y) = 19

P(X-Y)>5\\\\=1-0.00029\\=0.9997

C) For a group of 3, average has std deviation = \frac{11}{\sqrt{3} } \\=6.351

P(\bar y >115)\\= P(z>\frac{-5}{6.351} \\=0.7856

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2 years ago
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Step-by-step explanation:

The circumscribed circle or <em>circumcircle</em> of a polygon is a circle that passes through all the vertices of the polygon. The center of the circle, the circumcenter, is equidistant from all of the polygon's vertices.

The center is found at the point of intersection between the perpendicular bisectors of any two (non-parallel) chords of the circle. That is, <em>the perpendicular bisectors of any two of the sides of the triangle joining the cities will intersect at the circumcenter</em>.

The method of locating the center of the circle this way is simple and effective.

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Answer:


Step-by-step explanation:

Given is an algebraic polynomial of degree 5.

g(x) = 3x^5-2x^4+9x^3-x^2+12\\

Here leading term is p=3 and constant term is q =12

Factors of p are ±1,±2,±3

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