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goldfiish [28.3K]
3 years ago
13

Find the volume of a right circular cone that has a height of 17.6 m and a base with a

Mathematics
1 answer:
Firlakuza [10]3 years ago
5 0

Answer:

1179m^3

Step-by-step explanation:

Given radius=8m

Height=17.6m

Volume of the cone=1/3πr^2h

1/3 πr^2h

=1/3×3.14×8×8×17.6

=1178.96

Nearest tenth is 1179

So volume is 1179m^3

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Write √ 3 x √ 6 in the form b √2 where b is an integer.
tigry1 [53]

Answer:

Step-by-step explanation:

The catch here is to break up sqrt(6) into 2 parts that use primes to define sqrt(6)

sqrt(3)* sqrt(6)

sqrt(6) can be broken up into sqrt(2*3) which equals sqrt(2)*sqrt(3)

sqrt(3)*sqrt(2)(sqrt(3)

sqrt(3)*sqrt(3)*sqrt(2)

3*sqrt(2)

5 0
3 years ago
HELPPPPP PLZZZZ ASAPPPPP
iris [78.8K]

Answer:

complete angle is 360°

360 - 39

329°

7 0
3 years ago
A typical cup of coffee contains about 100 mg of caffeine and every hour approximately 16% ofthe amount of caffeine inthe body i
Over [174]

Answer:

a) \frac{dC}{dt} = rC

And for this case we can rewrite the model like this:

\frac{dC}{C} = r dt

If we integrate both sides we got:

ln C = rt + k

If we use exponentials for both sides we got:

C = e^{rt} e^k = C_o e^{rt}

For this case C_o = 100 mg and r = -0.16

So then our model would be given by:

C(t) = 100 e^{-0.16 t}

Where t represent the number of hours

b) C(5) = 100 e^{-0.16*5}= 44.9329

Step-by-step explanation:

Part a

For this case we can assume the proportional model given by:

\frac{dC}{dt} = rC

And for this case we can rewrite the model like this:

\frac{dC}{C} = r dt

If we integrate both sides we got:

ln C = rt + k

If we use exponentials for both sides we got:

C = e^{rt} e^k = C_o e^{rt}

For this case C_o = 100 mg and r = -0.16

So then our model would be given by:

C(t) = 100 e^{-0.16 t}

Where t represent the number of hours

Part b

For this case we can replace the value t=5 into the model and we got:

C(5) = 100 e^{-0.16*5}= 44.9329

7 0
3 years ago
Divide. Give special attention to the order of division. a. 72 ÷ 9 ÷ 2 b. (18 ÷ 6) ÷ 3 c. 45 ÷ 5 ÷ 3 d. 144 ÷ (12 ÷ 2)
Shkiper50 [21]
A. 72÷9÷2 = 8÷2 = 4
b. (18 ÷ 6) ÷ 3 = 3 ÷ 3 = 1
c. 45 ÷ 5 ÷ 3 = 9 ÷ 3 = 3
d. 144 ÷ (12 ÷ 2) = 144 ÷6 = 24
4 0
3 years ago
Read 2 more answers
Average Temperatures Suppose the temperature (degrees F) in a river at a point x meters downstream from a factory that is discha
mr Goodwill [35]

Answer:

Step-by-step explanation:

Average Temperatures Suppose the temperature (degrees F) in a river at a point x meters downstream from a factory that is discharging hot water into the river is given by

T(x) = 160-0.05x^2

a. [0, 10]

For x = 0

T(0) = 160 - 0.05 × 0^2

T(0) = 160

For x = 10

T(10) = 160 - 0.05 × 10^2

T(10) = 160 - 5 = 155

The average temperature

= (160 + 155)/2 = 157.5

b. [10, 40]

For x = 10

T(10) = 160 - 0.05 × 10^2

T(10) = 160 - 5 = 155

For x = 40

T(10) = 160 - 0.05 × 40^2

T(10) = 160 - 80 = 80

The average temperature

= (80 + 155)/2 = 117.5

c. [0, 40]

For x = 0

T(0) = 160 - 0.05 × 0^2

T(0) = 160

For x = 40

T(10) = 160 - 0.05 × 40^2

T(10) = 160 - 80 = 80

The average temperature

= (160 + 80)/2 = 120

6 0
3 years ago
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