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postnew [5]
4 years ago
13

im stuck on this . and somebody give me a clear right answer because people r giving me wrong answers

Mathematics
2 answers:
statuscvo [17]4 years ago
8 0

Answer:

-1

Step-by-step explanation:

S = <u>y</u> = <u>8</u> = -1

x 8

_______________________________________

Zarrin [17]4 years ago
7 0

Answer:

Slope = -1

Step-by-step explanation:

Take two points- the most obvious ones are (0,2) and (2,0)

Slope=(y2-y1)/(x2-x1)

=(2-0)/(0-2)

=2/-2

=-1

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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:

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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.

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3 years ago
I NEED THE ANSWR ASAP: Write the number as a power with a base of 5.<br> 125^3
EastWind [94]

Answer:

5^9

Step-by-step explanation:

I had this question before.

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3 years ago
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