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3 years ago

14

8. A cone shaped paper cup has a radius of 1.5 inches and a height of 3 inches. What is the volume of the cup?

1 answer:

Firdavs [7]3 years ago

6 0

Answer:

$V=7.06\ \text{inches}^3$

Step-by-step explanation:

Given hat,

The radius of cone shaped paper cup, r = 1.5 inches

The height of cone shaped paper cup, h = 3 inches

We need to find the volume of the cup. For a cone shaped, the volume is given by :

$V=\dfrac{1}{3}\pi r^2 h$

Putting all the values in the above formula.

$V=\dfrac{1}{3}\pi \times (1.5)^2 \times 3\\\\V=7.06\ \text{inches}^3$

So, the volume of the cup is $7.06\ \text{inches}^3$ .

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If you would like to find the inequality to determine the number of full-time employees Tom must schedule, you can calculate this using the following steps:

n ... the number of full-time employees

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The correct result would be at least 6 full-time employees.

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3 years ago

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Answer:

The correct option is D) $f(x) = x^3 + 2x^2 - 2x - 4$ .

Step-by-step explanation:

Consider the provided cubic function.

We need to find the equation having zeros: Square root of two, negative Square root of two, and -2.

A "zero" of a given function is an input value that produces an output of 0.

Substitute the value of zeros in the provided options to check.

Substitute x=-2 in $f(x) = x^3 + 2x^2 - 2x + 4$ .

$f(x) = x^3 + 2x^2 - 2x + 4\\f(x) = (-2)^3 + 2(-2)^2 - 2(-2) + 4\\f(x) =-8 + 2(4)+4 + 4\\f(x) =8$

Therefore, the option is incorrect.

Substitute x=-2 in $f(x) = x^3 + 2x^2 + 2x - 4$ .

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Therefore, the option is incorrect.

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$f(x) = x^3 - 2x^2 - 2x - 4\\f(x) = (-2)^3 - 2(-2)^2 - 2(-2) - 4\\f(x) =-8-8+4-4\\f(x) =-16$

Therefore, the option is incorrect.

Substitute x=-2 in $f(x) = x^3 + 2x^2 - 2x - 4$ .

$f(x) = x^3 + 2x^2 - 2x - 4\\f(x) = (-2)^3+2(-2)^2 - 2(-2) - 4\\f(x) =-8+8+4-4\\f(x) =0$

Now check for other roots as well.

Substitute x=√2 in $f(x) = x^3 + 2x^2 - 2x - 4$ .

$f(x) = x^3 + 2x^2 - 2x - 4\\f(x) = (\sqrt{2})^3+2(\sqrt{2})^2 - 2(\sqrt{2}) - 4\\f(x) =2\sqrt{2}+4-2\sqrt{2}-4\\f(x) =0$

Substitute x=-√2 in $f(x) = x^3 + 2x^2 - 2x - 4$ .

$f(x) = x^3 + 2x^2 - 2x - 4\\f(x) = (-\sqrt{2})^3+2(-\sqrt{2})^2 - 2(-\sqrt{2}) - 4\\f(x) =-2\sqrt{2}+4+2\sqrt{2}-4\\f(x) =0$

Therefore, the option is correct.

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Step-by-step explanation:

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