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3 years ago

Triangle KLM was dilated according to the rule

Mathematics

2 answers:

Vsevolod [243]3 years ago

7 0

Answer:

Correct choices are A, B and D

Step-by-step explanation:

The rule DO,0.75 (x,y) is a dilation by a scale factor of 0.75 about the origin. This rule can be rewritten as

(x,y)→(0.75x,0.75y) (option A is true).

The dilation preserves parallel lines, then LM is parallel to L'M' (option B is true).

According to the rule of dilation, $K'M'=0.75KM,$ this means that K'M' is shorter than KM (option C is false).

The vertices of the image are placed on the segments that connect origin with vertices of the pre-image triangle, then the vertices of the image are closer to the origin than those of the pre-image (option D is true).

The distance from M' to the origin is exactly 0.75 times the distance from M to the origin (option E is false).

MakcuM [25]3 years ago

4 0

The answer is
A. DO, 0.75 (x,y) = (0.75x, 0.75y)
B. LM is parallel to L'M'.
and C. The vertices of the image are closer to the origin than those of the pre-image.

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Answer:

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Step-by-step explanation:

a) what is the common difference of the sequence xn

Looking at the table, we get x_3=16, x_4=36 and x_5= 56

Deterring the common difference by subtracting x_4 from x_3 we get

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$x_n=x_1+(n-1)d\\x_{12}=x_1+(12-1)d\\x_{12}=-25+11(20)\\x_{12}=-25+220\\x_{12}=195$

So, $\mathbf{x_{12}=195}$

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Deterring the common difference by subtracting a_7 from a_8 we get

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We know common difference d= -13, we need to find $a_1$

Using $a_7=104$ we can find $x_1$

$a_n=a_1+(n-1)d\\a_7=a_1+(7-1)d\\104=a_1+7(-13)\\104=a_1-91\\a_1=104+91\\a_1=195$

So, We have $a_1 = 195$

Now finding $a_{12}$ , put n=12

$a_n=a_1+(n-1)d\\a_{12}=a_1+(12-1)d\\a_{12}=195+11(-13)\\a_{12}=195-143\\a_{12}=52$

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$a_n=a_1+(n-1)d\\a_{15}=a_1+(15-1)d\\a_{15}=195+14(-13)\\a_{15}=195-182\\a_{15}=13$

So, $\mathbf{a_{15}=13}$

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