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2 years ago

Need answer for d with working

Mathematics

1 answer:

son4ous [18]2 years ago

4 0

Answer:

Question D.

Pretty simple if you know what the distributive property is. In this equation, you notice 2 fractions outside the parenthesis. This means you must multiply this fraction by every number within its corresponding parenthesis. So, if you distribute the new equation would be $(\frac{5}{4}y+ 1)=(\frac{2}{3}y-\frac{1}{3})$ . After this isolate the variable y and solve. I'll leave that to you so you can get something out of it, because my entire explanation would be useless if you just looked at the answer and left, not knowing how to do it in the future. :)

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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=7 $e^{x}$

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

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

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Answer:

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Step-by-step explanation:

If x = 1

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