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-BARSIC- [3]
3 years ago
10

What are the square root of

t=" \sqrt{48} " align="absmiddle" class="latex-formula"> help mee please ​
Mathematics
2 answers:
PtichkaEL [24]3 years ago
6 0

<h3>Answer-</h3>

<h3>4√3 </h3>

<h3>Step by step explanation-</h3>

=\sqrt{2 \times 2 \times 2 \times 2 \times 3}

=\sqrt{2 {}^{2}  \times 2 {}^{2}  \times 3}

=2 \times 2 \sqrt{3}

= 4√3

<h3>So,The Answer is 4√3 .</h3>
Ray Of Light [21]3 years ago
5 0

\sqrt{48}  \\  =  \sqrt{2 \times 2 \times 2 \times 2 \times 3}  \\  =  \sqrt{ {2}^{2}  \times  {2}^{2}  \times 3}  \\  = 2 \times 2 \sqrt{3}  \\  = 4 \sqrt{3}

<h3>Your answer is 4√3.</h3>
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Read 2 more answers
Sistema de ecuaciones.5x+2y=-152x-2y=-6
Zolol [24]

Tnemos el sisema de ecuaciones:

\begin{gathered} 5x+2y=-15 \\ 2x-2y=-6 \end{gathered}

Podemos resolverlo por eliminación sumando ambas ecuaciones y eliminando y. Asi podemos resolver para x:

\begin{gathered} (5x+2y)+(2x-2y)=(-15)+(-6) \\ 7x+0y=-21 \\ x=-\frac{21}{7} \\ x=-3 \end{gathered}

Ahora podemos resolver para y con cualquiera de las dos ecuaciones:

\begin{gathered} 2x-2y=-6 \\ 2\cdot(-3)-2y=-6 \\ -6-2y=-6 \\ -2y=-6+6 \\ -2y=0 \\ y=0 \end{gathered}

Respuesta: x=-3, y=0

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1 year ago
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