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Stella [2.4K]
3 years ago
15

What is the radius of the cylinder please help

Mathematics
2 answers:
den301095 [7]3 years ago
8 0

Answer: it is 5

Step-by-step explanation:

Ipatiy [6.2K]3 years ago
4 0

Answer:

5cm

Step-by-step explanation:

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Emma used 1/7 of a liter of milk to make 1/3 of a jug of tea. How much milk, in liters, is required to fill the jug?
fiasKO [112]

Answer:

Step-by-step explanation:

Well....if 1/7 liters  of milk makes 1/6 of a jug

Then  six times this must be a full jug....so

6*(1/7)   =

6/7 liters of milk  required

i hope this helps you happy holidays let me know if you need more information

6 0
3 years ago
Read 2 more answers
Identify the linear term in the function of y = 2x2 − 5x − 12
MakcuM [25]

Answer:

The linear term is -5x

Step-by-step explanation:

The linear term is the term that has the degree equal to 1.

The given function is

y = 2x^2 − 5x − 12

here 2x^2 = quadratic term i.e having degree 2

-5x = linear term i.e having degree 1

-12 = constant

so, The linear term is -5x

4 0
3 years ago
I need help please
Dima020 [189]

It's either A or D.

6 0
3 years ago
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Suppose brine containing 0.2 kg of salt per liter runs into a tank initially filled with 500 L of water containing 5 kg of salt.
Oliga [24]

Answer:

(a) 0.288 kg/liter

(b) 0.061408 kg/liter

Step-by-step explanation:

(a) The mass of salt entering the tank per minute, x = 0.2 kg/L × 5 L/minute = 1 kg/minute

The mass of salt exiting the tank per minute = 5 × (5 + x)/500

The increase per minute, Δ/dt, in the mass of salt in the tank is given as follows;

Δ/dt = x - 5 × (5 + x)/500

The increase, in mass, Δ, after an increase in time, dt, is therefore;

Δ = (x - 5 × (5 + x)/500)·dt

Integrating with a graphing calculator, with limits 0, 10, gives;

Δ = (99·x - 5)/10

Substituting x = 1 gives

(99 × 1 - 5)/10 = 9.4 kg

The concentration of the salt and water in the tank after 10 minutes = (Initial mass of salt in the tank + Increase in the mass of the salt in the tank)/(Volume of the tank)

∴ The concentration of the salt and water in the tank after 10 minutes =  (5 + 9.4)/500 = (14.4)/500 = 0.288

The concentration of the salt and water in the tank after 10 minutes = 0.288 kg/liter

(b) With the added leak, we now have;

Δ/dt = x - 6 × (14.4 + x)/500

Δ = x - 6 × (14.4 + x)/500·dt

Integrating with a graphing calculator, with limits 0, 20, gives;

Δ = 19.76·x -3.456 = 16.304

Where x = 1

The increase in mass after an increase in = 16.304 kg

The total mass = 16.304 + 14.4 = 30.704 kg

The concentration of the salt in the tank then becomes;

Concentration = 30.704/500 = 0.061408 kg/liter.

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3 years ago
Help me please!!!!!!!
Komok [63]
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