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Maksim231197 [3]
2 years ago
9

Helpp me plssssss????

Mathematics
1 answer:
snow_tiger [21]2 years ago
3 0

Answer:

C

Step-by-step explanation:

The radius is 4 making the diameter 16 you need the other side since in half so that leaves you with answer C...16\pi yd^2

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T what point does the curve have maximum curvature? Y = 7ex (x, y) = what happens to the curvature as x → ∞? Κ(x) approaches as
Nookie1986 [14]

Formula for curvature for a well behaved curve y=f(x) is


K(x)= \frac{|{y}''|}{[1+{y}'^2]^\frac{3}{2}}


The given curve is y=7e^{x}


{y}''=7e^{x}\\ {y}'=7e^{x}


k(x)=\frac{7e^{x}}{[{1+(7e^{x})^2}]^\frac{3}{2}}


{k(x)}'=\frac{7(e^x)(1+49e^{2x})(49e^{2x}-\frac{1}{2})}{[1+49e^{2x}]^{3}}

For Maxima or Minima

{k(x)}'=0

7(e^x)(1+49e^{2x})(98e^{2x}-1)=0

→e^{x}=0∨ 1+49e^{2x}=0∨98e^{2x}-1=0

e^{x}=0  ,  ∧ 1+49e^{2x}=0   [not possible ∵there exists no value of x satisfying these equation]

→98e^{2x}-1=0

Solving this we get

x= -\frac{1}{2}\ln{98}

As you will evaluate {k(x})}''<0 at x=-\frac{1}{2}\ln98

So this is the point of Maxima. we get y=7×1/√98=1/√2

(x,y)=[-\frac{1}{2}\ln98,1/√2]

k(x)=\lim_{x\to\infty } \frac{7e^{x}}{[{1+(7e^{x})^2}]^\frac{3}{2}}

k(x)=\frac{7}{\infty}

k(x)=0







5 0
3 years ago
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Step-by-step explanation:

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14) y = x – 4<br> 4<br> A) 1<br> B)<br> 3<br> C)<br> D) -1<br> 3
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the answer is a

Step-by-step explanation:

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I need accurate answer please because this is the last attempt
matrenka [14]

Answer:

I'm not very sure, however I think you have to replace the sign - for a +.

Step-by-step explanation:

7 + 4√13

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