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Brrunno [24]
1 year ago
10

Lim x-> 0 sqrt x+4 -2 / x

Mathematics
1 answer:
Wittaler [7]1 year ago
3 0

Greetings from Brasil...

Here we have an indeterminacy 0/0

We can change variable or rationalize.......

Let's rationalize

{[√(X + 4) - 2]/X} · {[√(X + 4) + 2]/[√(X + 4) + 2]} = 1/[√(X + 4) + 2]

So, the limit will be 1/4

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If Olga was G years old two years ago, how old is she now? How old will she be in five years?
balandron [24]

Answer:

G+2 present age of Olga

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hope it helps.

8 0
3 years ago
Find the interval(s) where the function is increasing and the interval(s) where it is decreasing. (Enter your answers using inte
Mashcka [7]

Answer:

f'(x) > 0 on (-\infty ,\frac{1}{e}) and f'(x)<0 on(\frac{1}{e},\infty)

Step-by-step explanation:

1) To find and interval where any given function is increasing, the first derivative of its function must be greater than zero:

f'(x)>0

To find its decreasing interval :

f'(x)

2) Then let's find the critical point of this function:

f'(x)=\frac{\mathrm{d} }{\mathrm{d} x}[6-2^{2x}]=\frac{\mathrm{d} }{\mathrm{d}x}[6]-\frac{\mathrm{d}}{\mathrm{d}x}[2^{2x}]=0-[ln(2)*2^{2x}*\frac{\mathrm{d}}{\mathrm{d}x}[2x]=-ln(2)*2^{2x}*2=-ln2*2^{2x+1\Rightarrow }f'(x)=-ln(2)*2^{2x}*2\\-ln(2)*2^{2x+1}=-2x^{2x}(ln(x)+1)=0

2.2 Solving for x this equation, this will lead us to one critical point since x'  is not defined for Real set, and x''=\frac{1}{e}≈0.37 for e≈2.72

x^{2x}=0 \ni \mathbb{R}\\(ln(x)+1)=0\Rightarrow ln(x)=-1\Rightarrow x=e^{-1}\Rightarrow x\approx 2.72^{-1}x\approx 0.37

3) Finally, check it out the critical point, i.e. f'(x) >0 and below f'(x)<0.

8 0
3 years ago
Solve each equation algebraically using if–then statements to justify each step <br><br>2(x + 4) = 8
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