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hram777 [196]
3 years ago
8

Is w^2-9w+81 a trinomial square

Mathematics
1 answer:
Yuri [45]3 years ago
8 0
No because it would be
if you knew that
ax^2+bx+c was a trinomial square, then the factored form would be
( \sqrt{a}x+ \sqrt{c})^{2} or ( \sqrt{a}x- \sqrt{c})^{2}
so
if w^2-9w+81 is perfect square, factored form is either
( \sqrt{1}x+ \sqrt{81})^{2} or ( \sqrt{1}x- \sqrt{81})^{2}  aka ( x+9)^{2} or  ( x-9)^{2}
expand them
(x+9)^2=x^2+18x+81, wron middle term

(x-9)^2=x^2-18x+81 wrong

so it is not






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tamaranim1 [39]
Hello :
<span>2x+4y=38  ...(1)
10x+3y=105....(2)
-10x-20y = - 190
10x+3y =105 
add : -17y = - 85
y = 5
subsct in (1) : 2x +20  =38
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3 years ago
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zloy xaker [14]

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Step-by-step explanation:

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2 years ago
Can you Solve. 3x^2+10x+3
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to solve in ax^2+bx+c form you must find
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the numbers are 9 and 1 so

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factor
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3 years ago
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allochka39001 [22]

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So its not exactly direct variation as the last values is slightly different from the first 3.


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