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3 years ago

11

A new restaurant specializes in making pizza cones. A large slice of pizza is made into a cone shape and then filled with cheese

, meat, or vegetables. The cone formed by the slice of pizza measures 5 inches tall and has a diameter of 3 inches. How many cubic inches of fillings can the cone hold?

2 answers:

bagirrra123 [75]3 years ago

7 0

Answer: This is what I got and I thought there would be a 0 at the end but I guess not.

Step-by-step explanation:

I am Lyosha [343]3 years ago

3 0

The volume of the cone is πr²h/3 where r=3/2 and h=5. So volume=15π/4=11.78 cu in.

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Answer:

$x = \displaystyle \frac{5 + \sqrt{17}}{2}$ .

Step-by-step explanation:

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Similarly, because $(x - 4)$ and $(2\, x - 1)$ are two other inputs to the logarithm function in the original equation, they should also be positive. Therefore, $x > 4$ .

Let $a$ and $b$ represent two positive numbers (that is: $a > 0$ and $b > 0$ .) The following are two properties of logarithm:

$\displaystyle \ln (a) + \ln(b) = \ln\left(a \cdot b\right)$ .

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Apply these two properties to rewrite the original equation.

Left-hand side of this equation:

$\begin{aligned}&\ln(3\, x) - \ln(x - 4)= \ln\left(\frac{3\, x}{x -4}\right)\end{aligned}$

Right-hand side of this equation:

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Equate these two expressions:

$\begin{aligned}\ln\left(\frac{3\, x}{x -4}\right) = \ln(3(2\, x - 1))\end{aligned}$ .

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$\displaystyle \frac{3\ x}{x - 4} = 3\, (2\, x - 1)$ .

Simplify and solve this equation for $x$ :

$x^2 - 5\, x + 2 = 0$ .

There are two real (but not rational) solutions to this quadratic equation: $\displaystyle \frac{5 + \sqrt{17}}{2}$ and $\displaystyle \frac{5 - \sqrt{17}}{2}$ .

However, the second solution, $\displaystyle \frac{5 - \sqrt{17}}{2}$ , is not suitable. The reason is that if $x = \displaystyle \frac{5 - \sqrt{17}}{2}$ , then $(x - 4)$ , one of the inputs to the logarithm function in the original equation, would be smaller than zero. That is not acceptable because the inputs to logarithm functions should be greater than zero.

The only solution that satisfies the requirements would be $\displaystyle \frac{5 + \sqrt{17}}{2}$ .

Therefore, $x = \displaystyle \frac{5 + \sqrt{17}}{2}$ .

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