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2 years ago

Find the volume of a pyramid with a square base where the area of the base is 8.3 m 2 and the height of the pyramid is 9.8 m

Mathematics

1 answer:

borishaifa [10]2 years ago

7 0

Answer:

245.56

Step-by-step explanation:

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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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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

5 less than Y translates to:

Y - 5

Under 20 translates to:

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Therefore your answer is

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