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3 years ago

I really need help with this one.

Mathematics

1 answer:

lord [1]3 years ago

6 0

So all you really need to do is to substitute the value of x into the equation so like the 1st one will look like this: $f(-3) = 2(-3) + 4 = -2 \\ f(-1) = 2(-1) + 4 = 2 \\ f(0) = 2(0) + 4 = 4 \\ f(2) = 2(2) + 4 = 8 \\ f(4) = 2(4) + 4 = 12$ Those would be the right column for the f(x) values
Now do the same for g(x) $g(-3) = (-3)^2+1 = 10 \\ g(-1) = (-1)^2 + 1 = 2 \\ g(0) = 0^2 + 1 = 1 \\ g(2) = 2^2 + 1 = 5 \\ g(4) = 4^2 + 1 = 17$
You will notice that when x = -1, g(x) = f(x)

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shtirl [24]

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Step-by-step explanation:

20- 14.72=5.28. then divide 5.28 by .16(cents per minute) to get 33 minutes

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3 years ago

A cube is numbered 1 to 6 on its faces what is the theoretical probability of getting a 5 ? What is the theoretical probability

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3 years ago

Please help

cricket20 [7]

Answer:

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Step-by-step explanation:

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2 years ago

F(x) = lnx/x

My name is Ann [436]

$f(x) = \frac{lnx}{x}$

(a) $f'(x) = \frac{d}{dx}[\frac{lnx}{x}]$
Using the quotient rule:
$f'(x) = \frac{\frac{1}{x} \cdot x - lnx}{x^{2}}$
$f'(x) = \frac{1 - lnx}{x^{2}}$

For maximum, f'(x) = 0;
$\frac{1 - lnx}{x} = 0$
$lnx = 1, x = e$

(b) <em>Deduce:
</em> $e^{x} \geq x^{e}, x > 0$
<em>
Soln:</em> Since x = e is the greatest value, then f(e) ≥ f(x) > f(0)
$\frac{ln(e)}{e} \geq \frac{lnx}{x}$
$\frac{1}{e} \geq \frac{lnx}{x}$ , since ln(e) is simply equal to 1

Now, since x > 0, then we don't have to worry about flipping the signs when multiplying by x.
$\frac{x}{e} \geq lnx$
$x \geq elnx$
$x \geq ln(x^{e})$

Taking the exponential to both sides will cancel with the natural logarithmic function in the right hand side to produce:
$e^{x} \geq e^{ln(x^{e}})$
$e^{x} \geq x^{e}$ , as required.

3 0

3 years ago

Can someone please help me?

storchak [24]

5lbs=8.25
/5 /5
1lbs=1.65

6 0

3 years ago