2 years ago

Please help me with this

Mathematics

1 answer:

motikmotik2 years ago

8 0

The answer is the 15

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Math thanks ik u like math

Mariulka [41]

Oh man maths hard ;( ...

6 0

3 years ago

Read 2 more answers

Substitute t=3 and t=5 to determine if the two expressions are equivalent.

motikmotik

Answer:

because both sides are equal when substituted by 3 & 5

4 0

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.