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Rudiy27
3 years ago
15

Find the missing side length. If necessary, round to the nearest tenth.

Mathematics
1 answer:
Sphinxa [80]3 years ago
7 0

Answer:

25

Step-by-step explanation:

Let's call the missing side (the hypotenuse) c.

c² = 7²+24² (according to the pythagorean theorem)

c= \sqrt{625}

c=25



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At what point does the curve have maximum curvature? Y = 4ex (x, y) = what happens to the curvature as x → ∞? Κ(x) approaches as
MAXImum [283]

<u>Answer-</u>

At x= \frac{1}{2304e^4-16e^2} the curve has maximum curvature.

<u>Solution-</u>

The formula for curvature =

K(x)=\frac{{y}''}{(1+({y}')^2)^{\frac{3}{2}}}

Here,

y=4e^{x}

Then,

{y}' = 4e^{x} \ and \ {y}''=4e^{x}

Putting the values,

K(x)=\frac{{4e^{x}}}{(1+(4e^{x})^2)^{\frac{3}{2}}} = \frac{{4e^{x}}}{(1+16e^{2x})^{\frac{3}{2}}}

Now, in order to get the max curvature value, we have to calculate the first derivative of this function and then to get where its value is max, we have to equate it to 0.

 {k}'(x) = \frac{(1+16e^{2x})^{\frac{3}{2} } (4e^{x})-(4e^{x})(\frac{3}{2}(1+e^{2x})^{\frac{1}{2}})(32e^{2x})}{(1+16e^{2x} )^{2}}

Now, equating this to 0

(1+16e^{2x})^{\frac{3}{2} } (4e^{x})-(4e^{x})(\frac{3}{2}(1+e^{2x})^{\frac{1}{2}})(32e^{2x}) =0

\Rightarrow (1+16e^{2x})^{\frac{3}{2}}-(\frac{3}{2}(1+e^{2x})^{\frac{1}{2}})(32e^{2x})

\Rightarrow (1+16e^{2x})^{\frac{3}{2}}=(\frac{3}{2}(1+e^{2x})^{\frac{1}{2}})(32e^{2x})

\Rightarrow (1+16e^{2x})^{\frac{1}{2}}=48e^{2x}

\Rightarrow (1+16e^{2x})}=48^2e^{2x}=2304e^{2x}

\Rightarrow 2304e^{2x}-16e^{2x}-1=0

Solving this eq,

we get x= \frac{1}{2304e^4-16e^2}

∴ At  x= \frac{1}{2304e^4-16e^2} the curvature is maximum.




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2 years ago
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djverab [1.8K]
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In 1970, the average length of a major league baseball game was 150 minutes compared to 180 minutes in 2018. Calculate the absol
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