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4 years ago

Evaluate. Can someone please help me on this?

Mathematics

2 answers:

topjm [15]4 years ago

7 0

You must simplfy 4^-2 over 4^-4 and that is 0.0625/ .00390625 = 16

Then you multiply 12 x 16 = 192

Hope this helps :)

bagirrra123 [75]4 years ago

3 0

= 12 x (4^2)
= 12 x 16
= 192

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Simplify the expression

Dafna11 [192]

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3 years ago

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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

#learnwithBrainly

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olya-2409 [2.1K]

Answer:

5 times 3 and 5 divide by 1/3 yield the same result - see below.

Step-by-step explanation:

When we divide by a fraction we invert it and then multiply so:

5 / 1/3

= 5 * 3/1

= 5 * 3.

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