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Makovka662 [10]

4 years ago

11

Adult tickets to the fall play cost $8 and student tickets cost $4. The drama class sold 30 more adult tickets than student tick

ets to the fall play. If the class collected $840 from ticket sales, how many adult tickets were sold?
The drama class sold _____ adult tickets.

1 answer:

nignag [31]4 years ago

8 0

The class sold 80 adult tickets.<span />

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ddd [48]

12 pages

Each side count as a page

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4 years ago

A coin is tossed two times. The set of outcomes is

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Answer:

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3 years ago

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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3 years ago

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arsen [322]

Answer:

The coordinates are (-8,7).

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3 years ago

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A stone is thrown straight up from the edge of a roof, 750 feet above the ground at a speed of 14 feet per second. A) Remembering that the acceleration due to gravity is -32 feet per second squared, how high is the stone 2 seconds later? B) At what time does the stone hit the ground? C) What is the velocity of the stone when it hits the ground?

Step-by-step explanation:

A stone is thrown straight up from the edge of a roof, 750 feet above the ground at a speed of 14 feet per second. A) Remembering that the acceleration due to gravity is -32 feet per second squared, how high is the stone 2 seconds later? B) At what time does the stone hit the ground? C) What is the velocity of the stone when it hits the ground?

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3 years ago