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3 years ago

What is the simplest form of 6 radical 8 plus 8 radical 2

Mathematics

1 answer:

arsen [322]3 years ago

8 0

Answer:

20 $\sqrt{2}$

Step-by-step explanation:

Using the rule of radicals

$\sqrt{a}$ × $\sqrt{b}$ ⇔ $\sqrt{ab}$

Simplify

$\sqrt{8}$ = $\sqrt{4(2)}$ = $\sqrt{4}$ × $\sqrt{2}$ = 2 $\sqrt{2}$

Thus

6 $\sqrt{8}$ + 8 $\sqrt{2}$

= 6(2 $\sqrt{2}$ ) + 8 $\sqrt{2}$

= 12 $\sqrt{2}$ + 8 $\sqrt{2}$

= 20 $\sqrt{2}$

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Answer:

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3 years ago

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2 years ago

Find all the zeros of f(x) help please

Licemer1 [7]

Since $f(6)=0$ , by the remainder theorem this means that x - 6 divides f(x) exactly. This means there are constants a, b, and c such that

$\dfrac{f(x)}{x-6} = ax^2 + bx + c$

Multiplying both sides by x - 6 gives

$f(x) = (x-6)(ax^2+bx+c) \\\\ 2x^3-19x^2+45x-18 = ax^3 + (b-6a)x^2 + (c-6b)x - 6c$

Then a, b, c satisfy

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So, we have

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The quadratic term can be factored as

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which leaves us with

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2 years ago

2. If f(x) = x – 5 and g(x) = x2 + 3, find f[g(-2)].

irina [24]

Answer:

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Step-by-step explanation:

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2 years ago

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