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Maru [420]
3 years ago
11

Solve the simultaneous equations: 5x+4y=5 2x+7y=29

Mathematics
1 answer:
OLga [1]3 years ago
3 0

Answer:

x=−3 and y=5

Step-by-step explanation:

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A parking lot holds 42 cars. There are 26 cars in the lot already. Which inequality can be solved to show all the numbers of car
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16 parking lots still left.
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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
If a bracelet takes 5/6 foot of string how much string will it take to make 12 bracelets
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Step-by-step explanation:

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