Ask question

Ask question

All categories

3 years ago

13

If the height of the cone, h, is 10 millimeters and the diameter of the base, d, is 30 millimeters, what is the volume of the co

ne?
2,250 π mm3

750 π mm3

500 π mm3

250 π mm3

1 answer:

nirvana33 [79]3 years ago

8 0

750 pi mm cubed
1\3 (pi) r^2 h

You might be interested in

I have no idea. Please I need just answers. Thank you.

igomit [66]

Answer:

Step-by-step explanation:

1km = 1000m

5a. 5.768km = 5768m = 57.7m

5b. 9.039km = 9039m = 90.4m

5c. 8.48km = 8480m = 84.8m

5d. 17.685km = 17685m = 17.7m

5e. 17.658km = 17658m = 17.7m

To round to the nearest 100m just divide by 100 and then round it off. like 57.68=57.7m because 8 is more than 5 so 6 becomes 7 to one decimal.

Rule is applied for all the other questions too.

3 0

3 years ago

The lengths of two sides of a triangle are 11 cm and 19 cm. Identify the range of possible lengths for the third side.

mamaluj [8]

Answer:

2nd option

Step-by-step explanation:

Given 2 sides of a triangle then the third side x is in the range

difference of 2 sides < x < sum of 2 sides , that is

19 - 11 < x < 19 + 11

8 < x < 30

5 0

3 years ago

Read 2 more answers

Margo borrows $2000, agreeing to pay it back with 7% annual interest after 17 months. How much interest

julia-pushkina [17]

Answer:

1000

Step-by-step explanation:

bc i dont know thats why hope it helped

7 0

3 years ago

|r|=-2 solve the equation and graph the solution if possible

S_A_V [24]

That bracket symbol, with an r in it, means "the absolute value of r".
So r's absolute value is -2 meaning r is two units from zero
r= 2

6 0

3 years ago

Show that d^2y/dx^2=-2x/y^5, if x^3 + y^3=1

aalyn [17]

Answer:

y³ + x³ = 1

First, differentiate the first time, term by term:

${3y^{2}.\frac{dy}{dx} + 3x^{2}} = 0 \\\\{3y^{2}.\frac{dy}{dx} = -3x^{2}} \\\\\frac{dy}{dx} = \frac{-3x^{2}}{3y^{2}} \\\\\frac{dy}{dx} = \frac{-x^{2}}{y^{2}}$

↑ we'll substitute this later (4th step onwards)

Differentiate the second time:

$3y^{2}.\frac{dy}{dx} + 3x^{2} = 0 \\\\3y^{2}.\frac{d^{2} y}{dx^{2}} + 6y(\frac{dy}{dx})^{2} + 6x = 0 \\\\3y^{2}.\frac{d^{2} y}{dx^{2}} + 6y(\frac{dy}{dx})^{2} = - 6x \\\\3y^{2}.\frac{d^{2} y}{dx^{2}} + 6y(\frac{-x^{2} }{y^{2} })^{2} = - 6x \\\\3y^{2}.\frac{d^{2} y}{dx^{2}} + 6y(\frac{x^{4} }{y^{4} }) = - 6x \\\\3y^{2}.\frac{d^{2} y}{dx^{2}} + \frac{6x^{4} }{y^{3} } = - 6x \\\\3y^{2}.\frac{d^{2} y}{dx^{2}} = - 6x - \frac{6x^{4} }{y^{3} } \\\\$

$3y^{2}.\frac{d^{2} y}{dx^{2}} = - \frac{- 6xy^{3} - 6x^{4} }{y^{3}} \\\\\frac{d^{2} y}{dx^{2}} = - \frac{- 6xy^{3} - 6x^{4} }{3y^{2}. y^{3}} \\\\\frac{d^{2} y}{dx^{2}} = - \frac{- 2xy^{3} - 2x^{4} }{y^{5}} \\\\\frac{d^{2} y}{dx^{2}} = - \frac{-2x (y^{3} + x^{3})}{y^{5}} \\\\\frac{d^{2} y}{dx^{2}} = - \frac{-2x (1)}{y^{5}} \\\\\frac{d^{2} y}{dx^{2}} = - \frac{-2x}{y^{5}}$

3 0

3 years ago