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givi [52]
2 years ago
12

Find the missing digits for the following tracking number 84200912█

Mathematics
1 answer:
Lapatulllka [165]2 years ago
4 0
Whaaaaaaaaaaaaaaaaaat
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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
If you roll a number cube 12 times, about how many times would you expect to roll a 5 or 6
Marrrta [24]
You have a 1/3 chance each time you role So 1/3 x 12 = 4 times that you would role a 5 or a 6
6 0
2 years ago
Increase 60 by 25% please
GuDViN [60]

Answer:

15

Step-by-step explanation:

6 0
2 years ago
Read 2 more answers
**POINTS**Find the values of X and y
Andrej [43]

Answer:

y=105 for x

y=75

Step-by-step explanation:

2y-30=180                          180 (straight line)     180-105=75

+30      +30

2y=210

/2     /2

7 0
2 years ago
Find Y When x=2 y=5x-9​
nadya68 [22]
5*2-9 = y

10-9= 1
y=1
6 0
3 years ago
Read 2 more answers
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