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mr Goodwill [35]
2 years ago
5

Factor. ​9x^2+42xy+49y^2​ Enter your answer in the box

Mathematics
1 answer:
Sonbull [250]2 years ago
6 0

\qquad \qquad\huge \underline{\boxed{\sf ᴀɴsweʀ}}

The factorized form of the given equation is ~

\boxed{ \sf(3x + 7y) {}^{2} }

Let's solve ~

\qquad \sf  \dashrightarrow \: 9 {x}^{2}  + 42xy + 49 {y}^{2}

\qquad \sf  \dashrightarrow \: (3x) {}^{2}  + 2(3x  \times 7y) + (7y) {}^{2}

Now, as we can see, an Identity is applied here ~

that is ;

\qquad \sf  \dashrightarrow \:  {a}^{2} + 2ab +  {b }^{2}   = (a + b) {}^{2}

So, let's use this identity in our next step, taking :

  • a = 3x

  • b = 7y

\qquad \sf  \dashrightarrow \: (3x + 7y) {}^{2}

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Step-by-step answer:

-1/5n+7=2

-1/5n+7-7=2-7

-1/5n=-5

-1/5n/-1/5=-5/-1/5

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Please help me Algebra 1
Oduvanchick [21]

Answer:

a.  \sqrt{x^n} = x^\frac{n}{2}

b. \sqrt{x^n}  = x^\frac{(n-1)}{2}\sqrt{x}

Step-by-step explanation:

a) When <em>n </em>is even, then it is divisible by 2. Because of this, you can write:

  • \sqrt{x^n} = (x^n)^\frac{1}{2}
  • \sqrt{x^n} = x^\frac{n}{2}

b) When <em>n </em>is odd, then <em>n - 1 </em> is even. This would make it divisible by 2, and there would be a remainder of 1, so we can write:

  • \sqrt{x^n} = (x^n^-^1^+^1) ^\frac{1}{2}
  • \sqrt{x^n} = (x^n^-^1 × x)^\frac{1}{2}
  • \sqrt{x^n} = x^\frac{(n-1)}{2} × x^\frac{1}{2}
  • \sqrt{x^n}  = x^\frac{(n-1)}{2}\sqrt{x}
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Step-by-step explanation:

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