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3 years ago

Please I’m kinda lost on this one

Mathematics

1 answer:

Pie3 years ago

4 0

Answer:

I think the answer is D, but i am not 100% sure

Step-by-step explanation:

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Please help with this I am completely stuck on it

vaieri [72.5K]

Answer:

$f(x)=\sqrt[3]{x-4} , g(x)=6x^{2}\textrm{ or }f(x)=\sqrt[3]{x},g(x)=6x^{2} -4$

Step-by-step explanation:

Given:

The function, $H(x)=\sqrt[3]{6x^{2}-4}$

Solution 1:

Let $f(x)=\sqrt[3]{x}$

If $f(g(x))=H(x)=\sqrt[3]{6x^{2}-4}$ , then,

$\sqrt[3]{g(x)} =\sqrt[3]{6x^{2}-4}\\g(x)=6x^{2}-4$

Solution 2:

Let $f(x)=\sqrt[3]{x-4}$ . Then,

$f(g(x))=H(x)=\sqrt[3]{6x^{2}-4}\\\sqrt[3]{g(x)-4}=\sqrt[3]{6x^{2}-4} \\g(x)-4=6x^{2}-4\\g(x)=6x^{2}$

Similarly, there can be many solutions.

7 0

3 years ago

Factor: 3.x2 + 7x + 2

iren [92.7K]

Answer:

x= -1/3 and -2

I have made it in above picture

5 0

3 years ago

How do you find total pension expense?

chubhunter [2.5K]

Answer:

Total periodic pension costs (TPPC) is equal to the contributions plus change in the pension liability during the year. Each period, the periodic pension cost is recognized in profit or loss (P&L) and/or in other comprehensive income (OCI).

7 0

2 years ago

Determine if the sequence converges: [ln(n)]^2 /n

Finger [1]

We will use l´Hopital´s rule for calculating limits involving indeterminate form (in this case: ∞ / ∞ ) using the derivative of the numerator and denominator: $\lim_{n \to \infty} \frac{ln^{2}n }{n} = \lim_{n \to \infty} \frac{2ln(n)* \frac{1}{n} }{1} = \lim_{n \to \infty} \frac{2ln(n)}{n}$ This is still form ∞/∞ and we will use the derivative again: $\lim_{n \to \infty} \frac{2ln(n)}{n} = \lim_{n \to \infty} \frac{ \frac{1}{n} }{1} = \lim_{n \to \infty} \frac{1}{n}$ =1/∞ = 0
The sequence converges.

8 0

3 years ago

Which fraction is equivalent to -(6/-7)<br> A) -(6/7)<br> B) (-6/7)<br> C) (6/-7)<br> D) (-6/-7)

Len [333]

Given fraction:
-(6/-7)
Apply plus/minus rule:
-(-a)=a and -(a)=-a
So -(6/-7) = -6/-7= 6/7
Option D is correct.

Answer: (-6/-7)

4 0

3 years ago

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