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2 years ago

The base diameter and the height of a cone are both

Mathematics

1 answer:

WITCHER [35]2 years ago

3 0

Answer:

11/ 42. *. X^ 2. * X = I/3 X A * h; if E = 1/3 A

Then the answer is EX

Step-by-step explanation:

Volume of cone= 1/3 * base area * height

= 1/3 * pi* (X/2)^2 * X

= 1/3. * 22/ 7 * X^3/4

= 22/ 21 * X^3 / 4

= 11/21 *. X^2/ 2. *. X

= 11/ 42. * X^3

= 11/ 42. *. X^ 2. * X

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12 teams are competing in the final round of the world cup soccer conpetition. In how many different ways can the soccer teams f

alex41 [277]

Answer:

I don't know if this is correct, but I'm pretty sure that there are 12 different possible outcomes.

Step-by-step explanation:

The team could either be in 1st, 2nd, 3rd, 4th, 5th, 6th, 7th, 8th, 9th, 10th, 11th, or 12th place. This makes 12 outcomes.

7 0

3 years ago

Read 2 more answers

Addison worked 7 more hours this week than last week. In total, she worked 61 hours this week and last week. Using the equation

s2008m [1.1K]

D. 27

2h+7=61
Subtract 7 from both sides
2h=54
Divide by 2
Answer is 27

4 0

3 years ago

Read 2 more answers

The 3 angles of a triangle have a sum of 90 degrees.

coldgirl [10]

Each is 30° because you do 90/3 is 30

8 0

3 years ago

A gumball has a diameter that is 66 mm. The diameter of the gumball's spherical hollow core is 58 mm. What is the volume of the

grigory [225]

Answer:

Volume of gumball without including its hollow core is 48347.6 cubic mm.

Step-by-step explanation:

Given:

Diameter of Gumball = 66 mm

Since radius is half of diameter.

Radius of gumball = $\frac{diameter}{2}=\frac{66}{2} =33 \ mm$

Now We will first find the Volume of Gumball.

To find the Volume of Gumball we will use volume of sphere which is given as;

Volume of Sphere = $\frac{4}{3}\pi r^3$

Now Volume of Gumball = $\frac{4}{3}\times3.14 \times (33)^3 = 150456.24 \ mm^3$

Also Given

Diameter of gumball's spherical hollow core = 58 mm

Since radius is half of diameter.

Radius of gumball's spherical hollow core = $\frac{diameter}{2}=\frac{58}{2} =29 \ mm$

Now We will find the Volume of gumball's spherical hollow core.

Volume of Sphere = $\frac{4}{3}\pi r^3$

So Volume of gumball's spherical hollow core = $\frac{4}{3}\times3.14 \times (29)^3 = 102108.61 \ mm^3$

Now We need to find volume of the gumball without including its hollow core.

So, To find volume of the gumball without including its hollow core we would Subtract Volume of gumball spherical hollow core from Volume of Gumball.

volume of the gumball without including its hollow core = Volume of Gumball - Volume of gumball's spherical hollow core = $150456.24\ mm^3 - 102108.61\ mm^3 = 48347.63\ mm^3$

Rounding to nearest tenth we get;

volume of the gumball without including its hollow core = $48347.6\ mm^3$

Hence Volume of gumball without including its hollow core is 48347.6 cubic mm.

8 0

3 years ago

A scientist has 100 milligrams of a radioactive element. The amount of radioactive element remaining after t days can be determi

grandymaker [24]

The mass of the first shipment at time t is
$m_{1}=100( \frac{1}{2} )^{ \frac{t}{10}}$

The mass of the second shipment at time t is
$m_{2}=100( \frac{1}{2} )^{ \frac{t-3}{10} }$

At time t, the ratio of m₁ to m₂ is
$\frac{m_{1}}{m_{2}} = \frac{100}{100}. \frac{(1/2)^{t/10}}{(1/2)^{(t-3)/10}} \\ = \frac{(1/2)^{t/10}}{(1/2)^{t/10}}. \frac{1}{(1/2)^{-3/10}} \\ = (1/2)^{3/10} \\ = 0.8123$

Therefore as a percentage,
$\frac{m_{1}}{m_{2}} =100*0.8123 = 81.23 \%$

Answer: B. 81.2%

4 0

3 years ago