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Rufina [12.5K]
2 years ago
11

A sphere has a radius that is 9 inches long. What is the volume of the sphere?

Mathematics
1 answer:
Varvara68 [4.7K]2 years ago
6 0

Answer:

given

radius (r) = 8 inch

now,

V=4/3πr^3

=4/3×π×9^3

=3053.62806in³

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5. (9r 3 + 5r 2 + 11r) + (-2r 3 + 9r - 8r 2)
ad-work [718]

Answer:

7r³ - 3r² + 20r

Step-by-step explanation:

(9r³ + 5r² + 11r) + (-2r³ + 9r - 8r²)

You can get rid of the parentheses since there is nothing to distribute.

9r³ + 5r² + 11r - 2r³ + 9r - 8r²

Combine like factors.

7r³ - 3r² + 20r

3 0
3 years ago
The perimeter of a triangle is 26 centimeters. Side a of the
boyakko [2]

Answer:

side A 14, side b is11

Step-by-step explanation:

4 0
3 years ago
Regular pyramid <br> B=<br> P=<br> H=<br> l=<br> LA=<br> SA=<br> V=
devlian [24]

Answer:

B (Base surface area) = 16 m^{2}

P (Perimeter of the base) = 16 m

H (Height) = 6 m

L (Not sure what L means. I assume it's the slant height??) =  6.32 m

LA (I assume lateral surface area) = 50.59 m^{2}

SA (I assume total surface area) = 66.59 m^{2}

V (volume) = 32 m^{3}

Step-by-step explanation:

Sorry if anything is wrong. It would be more helpful to put the names, instead of acronyms.

7 0
3 years ago
Given that cos 57° = 0.545, enter the sine of a complementary angle.<br> sin<br> h
noname [10]

Answer:

sin 33° = 0.545

Step-by-step explanation:

complementary angle to 57° is 33°

sin 33° = 0.545

4 0
3 years ago
An urn contains three red balls and four blue balls. Draw two balls at random from the urn, without replacement. Compute the exp
mezya [45]

Step-by-step explanation:

in total we have 3+4 = 7 balls.

when we draw the first ball, the probability to draw a red ball is 3/7, and a blue ball 4/7.

when we draw the second ball, we have now only 6 balls in total.

the probabilty to draw a red back now depends also on the result of the first draw.

if the first ball was already red, then we have only a chance now of 2 out of 6.

if the first ball was blue, then we have now a chance of 3 out of 6.

so, the probably to draw at least 1 red ball in 2 draws is the probability of drawing one on the first draw plus the probability of drawing one on the second :

1 - probability to see 2 blue balls

1 - 4/7 × 3/6 = 1 - 12/42 = 30/42 = 0.714285714...

the expected number of red balls in 2 draws is

1 red in first red in first red in second

but not second and second but not first

1×(3/7 × 4/6) + 2×(3/7 × 2/6) + 1×(4/7 × 3/6) = 12/42 + 12/42 + 12/42 = 36/42 = 6/7 = 0.857142857 ≈ 0.8571

7 0
2 years ago
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