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2 years ago

Find the volume of a cone with a base radius of 5 in. and a height of 12 in.

Mathematics

2 answers:

Varvara68 [4.7K]2 years ago

5 0

Answer:

V = 314

Step-by-step explanation:

Volume of a cone:

V = πr²(h/3)

Given:

r = 5

h = 12

Work:

V = πr²(h/3)

V = 3.14(5²)(12/3)

V = 3.14(25)(4)

V = 314

Alona [7]2 years ago

5 0

Answer:

The volume of the cone is 311.1in³.

Step-by-step explanation:

V = ⅓Bh where B = πr²

V = ⅓πr²h

V = ⅓(22/7)(5²)12

V = 0.33(3.142)(25)12

V = 311.1in³

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drek231 [11]

An expression for the total cost of the daycare for Lucy's dog this week is T = 3F + 2H and the total cost is equal to $114.

<h3>How to write an expression for the total cost?</h3>

In this exercise, you're required to write an expression for the total cost of the daycare for Lucy's dog this week. Thus, we would assign a variable to the cost for full-day and half-day day care as follows:

Let the cost for full-day be F.

Let the cost for half-day be H.

Let total cost be T.

Thus, the total cost of the daycare is given by:

T = 3F + 2H

Evaluating the expression, we have:

T = 3(28) + 2(15)

T = 84 + 30

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Read more on expressions here: brainly.com/question/12189823

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Which function is undefined for x = 0? y=3√x-2 y=√x-2 y=3√x+2 y=√x=2

Andre45 [30]

For this case, we must indicate which of the given functions is not defined for $x = 0$

By definition, we know that:

$f (x) = \sqrt {x}$ has a domain from 0 to infinity.

Adding or removing numbers to the variable within the root implies a translation of the function vertically or horizontally. For it to be defined, the term within the root must be positive.

Thus, we observe that:

$y = \sqrt {x-2}$ is not defined, the term inside the root is negative when $x = 0$ .

While $y = \sqrt {x + 2}$ if it is defined for $x = 0.$

$f(x)=\sqrt[3]{x}$ , your domain is given by all real numbers.

Adding or removing numbers to the variable within the root implies a translation of the function vertically or horizontally. In the same way, its domain will be given by the real numbers, independently of the sign of the term inside the root.

So, we have:

$y = \sqrt [3] {x-2}$ with x = 0: $y = \sqrt [3] {- 2}$ is defined.