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3 years ago

Please help solve this

Mathematics

1 answer:

bazaltina [42]3 years ago

3 0

I cant fu(king read it.

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Liability that arises from not maintaining a building is referred to as a) medical foundation liability. b) premises liability.

Sauron [17]

Answer:

opt. B

Step-by-step explanation:

8 0

3 years ago

PLEASE HELP

bearhunter [10]

Answer:

i'm pretty sure 1.96×10-3 m³

is the right answer

7 0

3 years ago

Read 2 more answers

The sphere below has a radius of 2.5 inches and an approximate volume of 65.42 cubic inches.

Stells [14]

Part a: The radius of the second sphere is 5 inches.

Part b: The volume of the second sphere is 523.33 in³

Part c; The radius of the third sphere is 1.875 inches.

Part d: The volume of the third sphere is 27.59 in³

Explanation:

Given that the radius of the sphere is 2.5 inches.

Part a: We need to determine the radius of the second sphere.

Given that the second sphere has twice the radius of the given sphere.

Radius of the second sphere = 2 × 2.5 = 5 inches

Thus, the radius of the second sphere is 5 inches.

Part b: we need to determine the volume of the second sphere.

The formula to find the volume of the sphere is given by

$V=\frac{4}{3} \pi r^3$

Substituting $\pi=3.14$ and $r=5$ , we get,

$V=\frac{4}{3} (3.14)(125)$

$V=\frac{1580}{3}$

$V=523.3333 \ in^3$

Rounding off to two decimal places, we have,

$V=523.33 \ in^3$

Thus, the volume of the second sphere is 523.33 in³

Part c: We need to determine the radius of the third sphere

Given that the third sphere has a diameter that is three-fourths of the diameter of the given sphere.

Hence, we have,

Diameter of the third sphere = $\frac{3}{4} (5)=3.75$

Radius of the third sphere = $\frac{3.75}{2} =1.875$

Thus, the radius of the third sphere is 1.875 inches

Part d: We need to determine the volume of the third sphere

The formula to find the volume of the sphere is given by

$V=\frac{4}{3} \pi r^3$

Substituting $\pi=3.14$ and $r=1.875$ , we get,

$V=\frac{4}{3} (3.14)(1.875)^3$

$V=\frac{4}{3} (3.14)(6.59)$

$V=27.5901 \ in^3$

Rounding off to two decimal places, we have,

$V=27.59 \ in^3$

Thus, the volume of the third sphere is 27.59 in³

4 0

3 years ago

If R is between A and D and<br> DR = 8 and DA = 15 then AR=

dolphi86 [110]

Answer:

Step-by-step explanation:

heres a model

A ----------R---------D

A ----?-----R----8---D

A----------15----------D

15 - 8 = 7

AR = 7

5 0

3 years ago

Please please help!! I don't know how to do this!!

nexus9112 [7]

Using cosine law and considering the angle ABC is 89 you can deduce that AC is 6.3
6<6.3
So the largest possible integer is 6

7 0

3 years ago