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DochEvi [55]
3 years ago
10

Triangle CDA is the image of triangle ABC after a 180º rotation around the midpoint of segment AC. Triangle ECB is the image of

triangle ABC after a 180º rotation around the midpoint of segment BC.Quadrilateral A B D E. Point C lies on D E. Segments A C and A B are drawn inside the quadrilateral and each have a point marked.
Explain why ABCD and ABEC are parallelograms.
Mathematics
1 answer:
andrew-mc [135]3 years ago
7 0

Answer:

(1,-1), (5,-4), (2,-6)

Step-by-step explanation:

Took the test, this was the given answer.

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Please help meeeeeeeeeeeeeeee
il63 [147K]

Answer:

D

Step-by-step explanation:

You can google it and see if I'm right.... but I'm pretty sure I am

8 0
3 years ago
Read 2 more answers
Length of the median CP
ehidna [41]

Given:

ΔABC \sim ΔDEF

To find:

The length of median CP

Solution:

In ΔABC,

AP = 12, BP = 12 and PC = 3x - 12

In ΔDEF,

DQ = 16, QE = 16 and FQ = 2x + 8

If two triangles are similar, then their median is proportional to the corresponding sides.

$\Rightarrow \frac{AP}{DQ} =\frac{PC}{FQ}

$\Rightarrow \frac{12}{16} =\frac{3x-12}{2x+8}

Do cross multiplication.

$\Rightarrow 12(2x+8)=16(3x-12)

$\Rightarrow 24x+96=48x-192

Add 192 on both sides.

$\Rightarrow 24x+96+192=48x-192+192

$\Rightarrow 24x+288=48x

Subtract 24x from both sides.

$\Rightarrow 24x+288- 24x=48x- 24x

$\Rightarrow 288=24x

Divide by 24 on both sides.

⇒ 12 = x

Substitute x = 12 in CP.

CP = 3(12) - 12

     = 36 - 12

     = 24

The length of median CP is 24.

7 0
4 years ago
Help me with 16,17,18,and 19 please simpl
mixas84 [53]

Answer:

Ques 16)

We have to simplify the expression:

\dfrac{t^2}{t^2+3t-18}-(\dfrac{5t}{t^2+3t-18}-\dfrac{t-3}{t^2+3t-18})\\   \\=\dfrac{t^2}{t^2+3t-18}-(\dfrac{4t+3}{t^2+3t-18})\\  \\=\dfrac{t^2-4t-3}{t^2+3t+18}

Ques 17)

\dfrac{3w^2+7w-7}{w^2+8w+15}+\dfrac{2w^2-9w+4}{(2w^2+9w-5)(w^2-w-12)}\\  \\=\dfrac{3w^2+7w-7}{w^2+8w+15}+\dfrac{(2w-1)(w-4)}{(2w-1)(w+5)(w+3)(w-4)}\\\\=\dfrac{3w^2+7w-7}{(w+3)(w+5)}+\dfrac{1}{(w+3)(w+5)}\\\\=\dfrac{3w^2+7w-7+1}{(w+3)(w+5)}\\\\=\dfrac{(3w-2)(w+3)}{(w+5)(w+3)}\\\\=\dfrac{3w-2}{w+5}

Ques 18)

Let the blank space be denoted by the quantity 'x'.

\dfrac{x}{12a^2+8a}+\dfrac{15a^2}{12a^2+8a}=\dfrac{7a}{3a+2}\\ \\\dfrac{x+15a^2}{12a^2+8a}=\dfrac{7a}{3a+2}\\\\=\dfrac{x+15a^2}{4a(3a+2)}=\dfrac{7a}{3a+2}\\\\=\dfrac{x+15a^2}{4a}=7a\\\\x+15a^2=28a^2\\\\x=28a^2-15a^2\\\\x=13a^2

Ques 19)

Let the missing quantity be denoted by 'x'.

\dfrac{p^2+7p+2}{p^2+5p-14}-\dfrac{x}{p^2+5p-14}=\dfrac{p-1}{p-2}\\ \\\dfrac{p^2+7p+2-x}{p^2+5p-14}=\dfrac{p-1}{p-2}\\\\\dfrac{p^2+7p+2-x}{(p-2)(p+7)}=\dfrac{p-1}{p-2}\\\\p^2+7p+2-x=(p-1)(p+7)\\\\p^2+7p+2-x=p^2+6p-7\\\\x=p+9


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3 years ago
Solve for W. What does W equal?
Nostrana [21]

Answer:

w=-11/6

Step-by-step explanation:

5 0
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Answer:

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Step-by-step explanation:

4 0
4 years ago
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