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2 years ago

What is the height of a cone whose volume is 16 cubic units and whose radius is 3 units?

Mathematics

1 answer:

erica [24]2 years ago

6 0

Answer:

Square the radius, and then divide the radius squared into the tripled volume. For this example, the radius is 2. The square of 2 is 4, and 300 divided by 4 is 75. Divide the amount calculated in Step 2 by pi, which is an unending math constant that begins 3.14, to calculate the cone's height.

Step-by-step explanation:

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In a football tournament, the Bees scored 9 less than three times as many points as the Hornets. The Wasps scored 28 more points

Elina [12.6K]

Given:

Bees scored 9 less than three times as many points as the Hornets.

Wasps scored 28 more points than the Hornets.

Together the three teams scored 184 points.

To find:

The required equation for this scenario.

Step-by-step explanation:

Let x represent the number of points scored by the Hornets.

Bees scored 9 less than three times as many points as the Hornets.

Bees score = 3x-9

Wasps scored 28 more points than the Hornets.

Wasps score = x+28

Together the three teams scored 184 points.

$(3x-9)+x+(x+28)=184$

Therefore, the required equation is $(3x-9)+x+(x+28)=184$ .

3 0

3 years ago

Need help will give thanks and brainliest

Lina20 [59]

Answer : y = 11x + 24

7 0

2 years ago

Read 2 more answers

When it was Tess' turn in the "Guess my number game, the gloves were off! She gave Davis the following 10 clues:

astra-53 [7]

The number is 999777888.

Step-by-step explanation:

Reaching to the answer by solving the clues one by one.

The number is a 9 digit whole number. So it has to be 0 or greater than 0

The number is even .So its last digit will be even.

Each of the digit appears exactly thrice. So there will be three digits used three times each in the number.

Each digit is greater than 6. So there will be no digits less than 7 in the number.

All the digits in each period are same. So a period of three adjacent digits will have same the same digit.for ex.777 is a period

The number is greater than 7. So the digit in the millions place and higher is 9.

The number is divisible by 3. So the sum of all the digits should be divisible by 3.

The number is divisible by 4. So the number formed by last two digits of the number should be divisible by 4.

Thousands digit is 7.

The number that satisfies all the given conditions is 999777888.

5 0

2 years ago

Use this graph of velocity vs. time for two objects to answer the question.

Lynna [10]

Using derivatives, it is found that the correct option is:

A. Object C has an acceleration that is greater than the acceleration for D.

The acceleration is the derivative of the velocity, given by change in velocity divided by change in time, that is:

$a = \frac{\Delta_v}{\Delta_t}$

In this problem:

The change in time for objects C and T is the same.
The change in velocity for object C is greater, thus, it has a greater acceleration, and the correct option is:

A. Object C has an acceleration that is greater than the acceleration for D.

A similar problem is given at brainly.com/question/14516604

6 0

2 years ago

Read 2 more answers

Which expression is equivalent to *picture attached*

DiKsa [7]

Answer:

The correct option is;

$4 \left (\dfrac{50 (50+1) (2\times 50+1)}{6} \right ) +3 \left (\dfrac{50(51) }{2} \right )$

Step-by-step explanation:

The given expression is presented as follows;

$\sum\limits _{n = 1}^{50}n\times \left (4\cdot n + 3 \right )$

Which can be expanded into the following form;

$\sum\limits _{n = 1}^{50} \left (4\cdot n^2 + 3 \cdot n\right ) = 4 \times \sum\limits _{n = 1}^{50} \left n^2 + 3 \times\sum\limits _{n = 1}^{50} n$

From which we have;

$\sum\limits _{k = 1}^{n} \left k^2 = \dfrac{n \times (n+1) \times(2n+1)}{6}$

$\sum\limits _{k = 1}^{n} \left k = \dfrac{n \times (n+1) }{2}$

Therefore, substituting the value of n = 50 we have;

$\sum\limits _{n = 1}^{50} \left k^2 = \dfrac{50 \times (50+1) \times(2\cdot 50+1)}{6}$

$\sum\limits _{k = 1}^{50} \left k = \dfrac{50 \times (50+1) }{2}$

Which gives;

$4 \times \sum\limits _{n = 1}^{50} \left n^2 = 4 \times \dfrac{n \times (n+1) \times(2n+1)}{6} = 4 \times \dfrac{50 \times (50+1) \times(2 \times 50+1)}{6}$

$3 \times\sum\limits _{n = 1}^{50} n = 3 \times \dfrac{n \times (n+1) }{2} = 3 \times \dfrac{50 \times (51) }{2}$

$\sum\limits _{n = 1}^{50}n\times \left (4\cdot n + 3 \right ) = 4 \times \dfrac{50 \times (50+1) \times(2\times 50+1)}{6} +3 \times \dfrac{50 \times (51) }{2}$

Therefore, we have;

$4 \left (\dfrac{50 (50+1) (2\times 50+1)}{6} \right ) +3 \left (\dfrac{50(51) }{2} \right )$ .

4 0

3 years ago