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topjm [15]
3 years ago
11

one moving company charges $800 plus $16 per hour another moving company charges $720 plus $21 an hour at what number of hours w

ill the charge by both companies do the same what is the charge
Mathematics
1 answer:
Gekata [30.6K]3 years ago
6 0

800 + 16x = 720 + 21x. Cancel out the 720 to get 80 + 16x = 21x. Subtract 16x from both sides to get 80 = 5x. Divide by 5 to get 16 = x, or x = 16. 16 hours. The charge is 800 + 16(16), = 720 + 21(16) which = 800 + 256 = 720 + 336 which = 1,056 = 1,056. Hope this helped!

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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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