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Gelneren [198K]
2 years ago
6

(x-1)^2+(2-x)^2-3(x-5)*(x-5)

Mathematics
1 answer:
Natali5045456 [20]2 years ago
8 0

Answer:

-x^2+24x-70

Step-by-step explanation:

(x-1)^2+(2-x)^2-3(x-5)*(x-5)\\=> (x - 1)^2 + (2 - x)^2 - 3(x - 5)^2\\=> x^2 - 2x + 1 + 4 - 4x + x^2 - 3(x - 5)^2\\=> x^2 - 2x + 5 - 4x + x^2 - 3(x - 5)^2\\=> x^2 - 6x + 5 + x^2 - 3(x - 5)^2\\=> 2x^2 - 6x+5 - 3(x - 5)^2\\=> 2x^2 - 6x+5 - 3(x^2 - 10x + 25)\\=> 2x^2 - 6x+5  - 3x^2+30x-75\\=> -x^2 - 6x + 5 + 30x - 75\\=> -x^2 + 24x + 5-75\\=> -x^2+24x-70

Formula used =

(a-b)^2 = a^2+ 2ab + b^2

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\large{ \tt{❃ \: EXPLANATION}} :

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\large{ \tt{❁ \: USING \: PYTHAGORAS \: THEOREM}} :

\large{ \tt{❊ \:  {h}^{2}  =  {p}^{2}  +  {b}^{2} }}

\large{ \tt{⇢ {p}^{2} +  {b}^{2}   =  {h}^{2} }}

\large{ \tt{⇢ \:  {b}^{2}  =  {h}^{2}  -  {p}^{2} }}

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\large{ \tt{⇢ \:  {b}^{2}  = 1369 - 144}}

\large{ \tt{ ⇢{b}^{2}  = 1225}}

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  • Now , We know - Tan θ= \tt{ \frac{perpendicular}{base} }. Just plug the values :

\large{ \tt{➝ \: Tan  \: \theta =  \frac{p}{b}  = \boxed{ \tt{  \frac{12}{35} }}}}

▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁

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Bye!!!

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