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lisov135 [29]
3 years ago
10

ASAP!!!! FUNCTIONS HELP

Mathematics
1 answer:
Paladinen [302]3 years ago
3 0

Answer:

<h2>             3 </h2>

Step-by-step explanation:

All but first for every single x gives single y so the second, third and fourth are function

At first for x=2 we have two y (2 or -2) so this isn't function.

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The actual length of side t is 0.045 cm. Use the scale drawing to find the actual side length of w. A) 0.06 cm B) 0.075 cm C) 0.
Tems11 [23]

Answer: The answer is B 0.075 cm


Step-by-step explanation:

Since the drawings is a scale of the original you must use a proportion to solve the desired side. 0.045/.9=w/1.5;w=0.075.

5 0
3 years ago
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Layla used 0.482 grams of salt in her experiment. Maurcie use 0.51 grams of salt. Who used the greater amount of salt?
Montano1993 [528]
All you do is look at the first number and see which is greater than you will get your answer glad to help btw if you cant get it it's Maurice
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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 exponent of 10? <br><br> (Thank you for helping!)
Lemur [1.5K]

Answer:

exponent is 0

Step-by-step explanation:

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2 years ago
Logan has 30 balloons two fifths of the balloons are purple how many of the balloons are purple ​
Lapatulllka [165]

Answer:

Step-by-step explanation:

30\left(\frac{2}{5}\right)=12

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