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timurjin [86]
4 years ago
15

Henry's water bottle has a capacity of 1 liter. What is the capacity of Henry's water bottle in milliliters?

Mathematics
2 answers:
Nikolay [14]4 years ago
7 0
Hello,

The answer is "1,000ml".

Reason:

1 liter=1,000ml

If you need anymore help feel free to ask me!

Hope this helps!

~Nonportrit
kondor19780726 [428]4 years ago
4 0
1 liter is equal to 1000 milliliters

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What are the solution
AleksandrR [38]

Answer:

B

Step-by-step explanation:

We can get two equations from the inequality:

3x+2>9\\3x+2>-9

We just need to simplify both equations to get our answers:

3x+2>9\\3x>7\\x>\frac{7}{3}

3x+2>-9\\3x>-11\\x>\frac{-11}{3}

The two answers we get are:

x\frac{7}{3}

Which is also B.

3 0
2 years ago
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katrin [286]

Answer:

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7 0
3 years ago
Write the inequality represented by the graph below.
Ipatiy [6.2K]

Answer:

y > -x + 3

Step-by-step explanation:

find the slope and y-intercept so you can create a linear equation:

y = mx + b ;   m = slope and  b = y-intercept

slope (m) = (3-0) / (0-3)

m = 3/-3 or -1

we can see the y-intercept by looking at the graph, it is 3

y > -x + 3

6 0
2 years ago
Suppose that a large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved. Another br
Lina20 [59]

Answer:

The differential equation for the amount of salt A(t) in the tank at a time  t > 0 is \frac{dA}{dt}=12 - \frac{2A(t)}{500+t}.

Step-by-step explanation:

We are given that a large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved. Another brine solution is pumped into the tank at a rate of 3 gal/min, and when the solution is well stirred, it is then pumped out at a slower rate of 2 gal/min.

The concentration of the solution entering is 4 lb/gal.

Firstly, as we know that the rate of change in the amount of salt with respect to time is given by;

\frac{dA}{dt}= \text{R}_i_n - \text{R}_o_u_t

where, \text{R}_i_n = concentration of salt in the inflow \times input rate of brine solution

and \text{R}_o_u_t = concentration of salt in the outflow \times outflow rate of brine solution

So, \text{R}_i_n = 4 lb/gal \times 3 gal/min = 12 lb/gal

Now, the rate of accumulation = Rate of input of solution - Rate of output of solution

                                                = 3 gal/min - 2 gal/min

                                                = 1 gal/min.

It is stated that a large mixing tank initially holds 500 gallons of water, so after t minutes it will hold (500 + t) gallons in the tank.

So, \text{R}_o_u_t = concentration of salt in the outflow \times outflow rate of brine solution

             = \frac{A(t)}{500+t} \text{ lb/gal } \times 2 \text{ gal/min} = \frac{2A(t)}{500+t} \text{ lb/min }

Now, the differential equation for the amount of salt A(t) in the tank at a time  t > 0 is given by;

= \frac{dA}{dt}=12\text{ lb/min } - \frac{2A(t)}{500+t} \text{ lb/min }

or \frac{dA}{dt}=12 - \frac{2A(t)}{500+t}.

4 0
3 years ago
What is the are of the figure below​
Vladimir [108]

Answer:

The answer would most likely to eight. If not use the formula Base x Height

Step-by-step explanation:

4 0
3 years ago
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