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3 years ago

Help with finding the volume of the cone pictured

Mathematics

1 answer:

jasenka [17]3 years ago

4 0

Volume of the cone = 56.52 in³

Solution:

Diameter of the cone = 6 in

Radius of the cone = 6 ÷ 2 = 3 in

Height of the cone = 6 in

The value of π = 3.14

<u>To find the volume of the cone:</u>

Volume of the cone = $\frac{1}{3} \pi r^{2} h$

$$=\frac{1}{3} \times 3.14 \times 3^{2} \times 6$

$$=3.14 \times 18$

= 56.52

Volume of the cone = 56.52 in³

Hence the volume of the given cone is 56.52 $\text{in}^3$ .

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In a recent semester at a local university, 490 students enrolled in both General Chemistry and Calculus I. Of these students, 6

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3 years ago

The depth of snow after n hours of a snowstorm is represented by the function f(n + 1) = f(n) + 0.8 where f(0) = 2.5. Which stat

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4 years ago

Read 2 more answers

9x^3+6x^2-3x Please!!! HELP!!! need it done asap

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3 years ago

How many pounds of a 15% copper alloy must be mixed with 700lb of a 30% copper alloy to maybe a 25.5% copper alloy

lisov135 [29]

Answer:

$300\; \rm lb$ .

Step-by-step explanation:

Let $x$ represent the mass (in pounds) of that $15\%$ copper alloy required, such that the final mixture would contain $25.5\%$ copper by mass.

Consider: if $x$ pounds of that $15\%$ copper alloy is mixed with $700$ pounds that $30\%$ copper alloy, what would be the mass of copper in the mixture?

Mass of copper in $x$ pounds of that $15\%$ copper alloy: $(0.15\, x)\; \rm lb$ .
Mass of copper in $700$ pounds of that $30\%$ copper alloy: $700 \times 0.30 = 210\; \rm lb$ .

Therefore, the mixture would contain $(210 + 0.15\, x) \; \rm lb$ of copper.

The mass of that mixture would be $(700 + x)\; \rm lb$ . The mass fraction of copper in that mixture would be:

$\displaystyle \frac{(210 + 0.15\, x)\; \rm lb}{(700 + x)\; \rm lb} \times 100\%$ .

This ratio is supposed to be equal to $25.5\%$ . These two pieces of equations combine to give an equation about $x$ :

$\displaystyle \frac{(210 + 0.15\, x)\; \rm lb}{(700 + x)\; \rm lb} \times 100\% = 25.5\%$ .

$\displaystyle \frac{210 + 0.15\, x}{700 + x} = 0.255$ .

Simplify and solve for $x$ :

$210 + 0.15\, x= 0.255\, (700 + x)$ .

$(0.255 - 0.15)\, x= 210 - 0.255 \times 700$ .

$\displaystyle x = \frac{210 - 0.255 \times 700}{0.255 - 0.15} = 300$ .

Therefore, $300\; \rm lb$ of that $15\%$ alloy would be required.

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3 years ago

Lim x--> 0 (e^x(sinx)(tax))/x^2

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Make use of the known limit,

$\displaystyle\lim_{x\to0}\frac{\sin x}x=1$

We have

$\displaystyle\lim_{x\to0}\frac{e^x\sin x\tan x}{x^2}=\left(\lim_{x\to0}\frac{e^x}{\cos x}\right)\left(\lim_{x\to0}\frac{\sin^2x}{x^2}\right)$

since $\tan x=\dfrac{\sin x}{\cos x}$ , and the limit of a product is the same as the product of limits.

$\dfrac{e^x}{\cos x}$ is continuous at $x=0$ , and $\dfrac{e^0}{\cos 0}=1$ . The remaining limit is also 1, since

$\displaystyle\lim_{x\to0}\frac{\sin^2x}{x^2}=\left(\lim_{x\to0}\frac{\sin x}x\right)^2=1^2=1$

so the overall limit is 1.

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3 years ago