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Ask QuestionPosted by Muskan Kumari 6 years, 11 months ago
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Yogita Ingle 6 years, 11 months ago
x = 2 , y = 3
2x + 3y - 13
= 2(2) + 3(3) - 13
= 4 + 9 - 13
= 13 - 13
= 0
Posted by Binder Kour 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
Let a be any positive integer and b = 6
∴ by Euclid’s division lemma
a = bq + r, 0≤ r and q be any integer q ≥ 0
∴ a = 6q + r,
where, r = 0, 1, 2, 3, 4, 5
If a is even then then remainder by division of 6 is 0,2 or 4
Hence r=0,2,or 4
or A is of form 6q,6q+2,6q+4
As, a = 6q = 2(3q), or
a = 6q + 2 = 2(3q + 1), or
a = 6q + 4 = 2(3q + 2).
If these 3 cases a is an even integer.
but if the remainder is 1,3 or 5 then r=1,3 or 5
or A is of form 6q+1,,6q+3 or,6q+5
Case 1:a = 6q + 1 = 2(3q) + 1 = 2n + 1,
Case 2: a = 6q + 3 = 6q + 2 + 1,
= 2(3q + 1) + 1 = 2n + 1,
Case 3: a = 6q + 5 = 6q + 4 + 1
= 2(3q + 2) + 1 = 2n + 1
This shows that odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.
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Given: ABCD is a parallelogram and E is a point on BC. The diagonal BD intersects AE at F.
To prove: {tex}A F \times F B = E F \times F D{/tex}
Proof: Since ABCD is a parallelogram, then its opposite sides must be parallel.
{tex}\therefore{/tex}In {tex}\triangle ADF{/tex} and {tex}\triangle EBF{/tex}
{tex}\angle \mathrm { FDA } = \angle \mathrm { EBF }{/tex} and {tex}\angle F A D = \angle F E B{/tex} [Alternate interior angles]
{tex}\angle A F D = \angle B F E{/tex} [vertically opposite angles]
{tex}{/tex} Therefore,by AAA criteria of similar triangles,we have,
{tex}\triangle \mathrm { ADF } = \triangle \mathrm { EBF }{/tex}
Since the corresponding sides of similar triangles are proportional. Therefore,we have,
{tex} \frac { A F } { F D } = \frac { E F } { F B }{/tex}
{tex}\Rightarrow A F \times F B = E F \times F D{/tex}

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Riya Trivedi 6 years, 11 months ago
2Thank You