Answer:
(4x + 7y)(4x - 7y)
Step-by-step explanation:
Rewrite 16 as 4^2
= 4^2x^2 - 49y^2
Rewrite 49 as 7^2
= 4^2x^2 - 7^2y^2
Apply the Exponent Rule Pt 1 ((a^m*b^m =(ab)^m))
= (4x)^2 - 7^2y^2
Apply the Exponent Rule Pt 2
= (4x)^2 - (7y)^2
Apply Difference of Squares Formula (( x^2-y^2 = (x + y)(x - y)
= (4x + 7y) (4x - 7y)
10mn -35m
8rs-2r^2+16r
they are in order.
r = 4/(x+y)
rx + ry = 4
Factor out r
r ( x+y) = 4
Divide each side by (x+y)
r ( x+y) /(x+y) = 4/(x+y)
<em>AC bisects ∠BAD, => ∠BAC=∠CAD ..... (1)</em>
<em>thus in ΔABC and ΔADC, ∠ABC=∠ADC (given), </em>
<em> ∠BAC=∠CAD [from (1)],</em>
<em>AC (opposite side side of ∠ABC) = AC (opposite side side of ∠ADC), the common side between ΔABC and ΔADC</em>
<em>Hence, by AAS axiom, ΔABC ≅ ΔADC,</em>
<em>Therefore, BC (opposite side side of ∠BAC) = DC (opposite side side of ∠CAD), since (1)</em>
<em />
Hence, BC=DC proved.
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