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3 years ago

Evaluate. 32−[(4+2×3)×2] Thk u!

Mathematics

2 answers:

gulaghasi [49]3 years ago

7 0

Answer:

$here \: w e \: can \: use \: bodmas \\ 32 - ((4 + 2 \times 3) \times 2) \\ = 32 - (4 + 6) \times 2) \\ = 32 - 10 \times 2 \\ = 32 - 20 \\ = 12 \\ thank \: yuou$

Citrus2011 [14]3 years ago

4 0

Answer:

Then the answer will be 32

Step-by-step explanation:

$32 - ((4 + 2 \times 3) \times 2) \\ 32 - ((4 + 6) \times 2) \\ 32 - (10 \times 2) \\ 32 - 10 \\ = 22$

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What are the roots of f(x) = x2 – 48? –48 and 48 –24 and 24 Negative 8 StartRoot 3 EndRoot and 8 StartRoot 3 EndRoot Negative 4

Ivanshal [37]

Answer:

$x=4\sqrt{3},\:x=-4\sqrt{3}$ are the roots.

Step-by-step explanation:

$x=4\sqrt{3},\:x=-4\sqrt{3}$

Considering the expression

$x^2\:-\:48$

Solving

$x^2-48=0$

$x^2-48+48=0+48$

$x^2=48$

$\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}$

$x=\sqrt{48},\:x=-\sqrt{48}$

Solving

$x=\sqrt{48}$

= $\sqrt{2^4\cdot \:3}$

$\mathrm{Apply\:radical\:rule}:\quad \sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b}$

= $\sqrt{3}\sqrt{2^4}$

$\mathrm{Apply\:radical\:rule}:\quad \sqrt[n]{a^m}=a^{\frac{m}{n}}$

$\sqrt{2^4}=2^{\frac{4}{2}}=2^2$

$=2^2\sqrt{3}$

$=4\sqrt{3}$

So,

$x=4\sqrt{3}$

Similarly,

$x=-\sqrt{48}=-4\sqrt{3}$

Therefore, $x=4\sqrt{3},\:x=-4\sqrt{3}$ are the roots.

Keywords: roots, expression

Learn more about roots from brainly.com/question/3731376

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