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3 years ago

Glven: ABCD is a parallelogram. Prove: AB - CD and BC – DA

Mathematics

2 answers:

coldgirl [10]3 years ago

7 0

Answer:

$by \: rotation : \\ { \bf{AB + BC = CD+ AD}} \\ { \boxed{ \tt{AB - CD = BC - DA}}}$

since: DA = -AD

Lena [83]3 years ago

3 0

Answer:

Here are all the answers (this took me awhile to figure out myself)

Step-by-step explanation:

Statement 1: ABCD is a parallelogram.

Reason 1: Given

Statement 2: draw AC.

Reason 2: Unique line postulate.

Statement 3: BCA & DAC are alt. interior angles.

Reason 3: def. of alt. interior angles.

Statement 4: DCA & BAC are alt. interior angles.

Reason 4: def. of alt. interior angles.

Statement 5: AB || CD

Reason 5: def. of parallelogram

Statement 6: BC || DA

Reason 6: def. of parallelogram

Statement 7: <BCA = <DAC

Reason 7: alternative interior angle theorem

Statement 8: <DCA = <BAC

Reason 8: alternative interior angle theorem

Statement 9: AC = AC

Reason 9: reflexive property

Statement 10: ABC = CDA

Reason 10: ASA

Statement 11: AB = CD

Reason 11: CPCTC

Statement 12: BC = DA

Reason 12: CPCTC

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Answer:

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Step-by-step explanation:

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The first two terms in an arithmetic progression are -2 and 5. The last term in the progression is the only number in the progre

Rudik [331]

Given:

The first two terms in an arithmetic progression are -2 and 5.

The last term in the progression is the only number in the progression that is greater than 200.

To find:

The sum of all the terms in the progression.

Solution:

We have,

First term : $a=-2$

Common difference : $d = 5 - (-2)$

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$a_n=a+(n-1)d$

where, a is first term and d is common difference.

$a_n=-2+(n-1)(7)$

According to the equation, $a_n>200$ .

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$(n-1)(7)>202$

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Add 1 on both sides.

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Sum of n terms of an A.P. is

$S_n=\dfrac{n}{2}[2a+(n-1)d]$

Substituting n=30, a=-2 and d=7, we get

$S_{30}=\dfrac{30}{2}[2(-2)+(30-1)7]$

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$S_{30}=15[-4+203]$

$S_{30}=15(199)$

$S_{30}=2985$

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Alenkasestr [34]

Answer:

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Step-by-step explanation:

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Answer:

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