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3 years ago

What's 121,580 rounded to the nearest thousand?

Mathematics

2 answers:

HACTEHA [7]3 years ago

6 0

When you round 121,580 to the nearest thousand, you get 122,000. Because 5 there so we round up.

iragen [17]3 years ago

4 0

The answer is 122,000

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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=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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Solution:

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=> x⁴ + 11x³ + 23x² - 95x - 300

The only term that has a x-variable is "-95x".

The coefficient of x is -95.

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