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siniylev [52]
3 years ago
5

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=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
(X^2+ x– 12) (x^2+ 10x + 25)
yulyashka [42]

Solution:

  • (x² + x – 12)(x² + 10x + 25)
  • => (x⁴ + 10x³ + 25x²) + (x³ + 10x² + 25x) + (-12x² - 120x - 300)
  • => x⁴ + 10x³ + 25x² + x³ + 10x² + 25x - 12x² - 120x - 300
  • => x⁴ + (10x³ + x³) + (25x² + 10x² - 12x²) + (25x - 120x) - 300
  • => x⁴ + (11x³) + (23x²) + (-95x) - 300
  • => x⁴ + 11x³ + 23x² - 95x - 300

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

The coefficient of x is -95.

6 0
2 years ago
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Answer:

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Answer:

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