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Greeley [361]
3 years ago
13

In the new logo, what fraction of the parallelogram is shaded?

Mathematics
1 answer:
Harman [31]3 years ago
7 0
D. 1/4 because imagine splitting that parallelogram in half, and you will see on the first half that half of that half is shaded. So that part is 1/2 and if you add the second half which isn't shaded at all, that's 0/2 so do 1/2 plus 0/2 and you will get 1/4
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What is the solution set for 6x−2=5x+3, given the replacement set {2, 3, 4, 5}?
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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}}

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{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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