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4 years ago

What is h(-2) H(-1.75) H(1)

Mathematics

1 answer:

ruslelena [56]4 years ago

3 0

$h(x)=\left\{\begin{array}{cccc}-2&if&-2\leq x$

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yKpoI14uk [10]

Answer: yes

Step-by-step explanation:

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2 years ago

Three water storage containers have the same volume in the shape of a cylinder, a cone, and a sphere. The base radius of the cyl

fredd [130]

Answer:

- The ratio of the height of the cone to the height of the cylinder is 3:1

- The ratio of the height of the cone to radius of the sphere is 4:1

Step-by-step explanation:

Note that the radii of all the structures are the same.

Let the height of the cone be H and the height of the cylinder be h

The volume of each of the shapes are given below as

Volume of cylinder = πr²h

Volume of a cone = (1/3)πr²H

Volume of a sphere = (4/3)πr³

1) The ratio of the height of the cone to the height of the cylinder

To obtain this, we equate the volume of those two structures

Volume of the cone = Volume of the cylinder

(1/3)πr²H = πr²h

πr² cancels out on both sides and we're left with

(H/3) = h

H = 3h

(H/h) = (3/1)

So, The ratio of the height of the cone to the height of the cylinder is 3:1

2) The ratio of the height of the cone to radius of the sphere

Similarly equating the volumes of the cone and the sphere

(1/3)πr²H = (4/3)πr³

(1/3)πr² cancels out on both sides and we're left with

H = 4r

(H/r) = (4/1)

The ratio of the height of the cone to radius of the sphere is 4:1

Hope this Helps!!!

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4 years ago

Use the exponent properties to simplify the following<br> <img src="https://tex.z-dn.net/?f=%5Cfrac%7B3x%5E3y%5E2z%7D%7B27xy%5E2

7nadin3 [17]

Step-by-step explanation:

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3 years ago

Negation of no planes have three wheels

anygoal [31]

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3 years ago

Seven blue and four red balls are to be arranged in order. How many ways can this be done if (1) The blue balls are distinguisha

Sav [38]

Answer:

1) n = 39916800

2) n = 1663200

3) n = 330

Step-by-step explanation:

1) If the blue balls are distinguishable as are the red balls

Then you can arrange these balls in the following ways, we must use a permutation

In totally we have 11 balls, then

n = 11P11

$n = \frac{11!}{(11-11)!} = 11! = 39916800\\$

2) If Blue balls are distinguishable, but the red balls are identical

In this case, we need to do a correction due to the red balls are identical and we cannot identify the difference when we interchange two red balls

$n = \frac{11!}{4!} = \frac{39916800}{24} = 1663200$

3) If the balls of each color are indistinguishable

We proceed equal to the before case but we include a correction due to blue balls also

$n = \frac{11!}{4!*7!} = \frac{39916800}{24*5040} = 330$

7 0

3 years ago