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3 years ago

What is the approximate volume of a right cone with a diameter of 10 meters and a height of 5 meters? Use 3.14 for π.

Mathematics

1 answer:

Nikolay [14]3 years ago

5 0

A. 130.8 m^3, you would use the formula for a cone it's: V= 1/3 x pi x r^2 x h

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Figure B is a scaled copy of Figure A.

elixir [45]

Answer:

The scale factor would be 1/2.

Step-by-step explanation:

2 of the sides both are 4 units.

If you go to figure B, it got smaller.

So in this case, you would divide.

On figure B, 2 of the sides are 2 units.

If you did it in reverse, you would get 2 because 2 times 2 is 4.

But we divide, so 4/2 is 2, so the scale factor would be:

1/2

Hope this helped!

And brainliest pls :)

6 0

2 years ago

PLease Help! I'm hopeless at math.

Furkat [3]

Answer:

It's B, I'm only trying my best since there is no picture.

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

What is the value of x for part A and part B

GuDViN [60]

Part A :

So the two sides are equivalent,

$9x - 18 = 3x$

Solve for x and get 3.

~~

Answer : x = 3

~~

Part B :

To find JM, simply substitute x for 3 and solve,

$9(3) - 18$
$27 - 18 = 9$
~^

Answer : JM = 9

~~

I hope that helps you out!!

Any more questions, please feel free to ask me and I will gladly help you out!!

~Zoey

5 0

3 years ago

Anyone trying to help a girl out?

adoni [48]

Answer:

so it is a rectangle with a perimeter of 30 units

3 0

3 years ago

Solve the following equation. X cubed minus 6X squared plus 6X equals zero

anyanavicka [17]

We have to solve this equation:

$x^3-6x^2+6x=0$

Third degree polynomials like this one are not easily solved, but this one has a root at x = 0. The let us factorize this polynomial as x times a second degree polynomial:

$\begin{gathered} x^3-6x^2+6x=0 \\ x(x^2-6x+6)=0 \end{gathered}$

Now we can find the roots of the quadratic polynomial as:

$\begin{gathered} x=\frac{-(-6)\pm\sqrt[]{(-6)^2-4\cdot1\cdot6}}{2\cdot1} \\ x=\frac{6\pm\sqrt[]{36-24}}{2} \\ x=\frac{6\pm\sqrt[]{12}}{2} \\ x=\frac{6\pm\sqrt[]{4\cdot3}}{2} \\ x=\frac{6\pm2\sqrt[]{3}}{2} \\ x=3\pm\sqrt[]{3} \\ x_1=3-\sqrt[]{3} \\ x_2=3+\sqrt[]{3} \end{gathered}$

Then, the solutions to the equation are:

x = 0

x = 3 - √3

x = 3 + √3

4 0

1 year ago