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Sophie [7]
3 years ago
14

Can ya'll please help? I'll give brainliest!!

Mathematics
2 answers:
zlopas [31]3 years ago
8 0

Answer:

Insightful, motivational, inspiring, helpful, useful, kind, caring, welcoming, lovely

Step-by-step explanation:

I hope this was what you were looking for

Schach [20]3 years ago
7 0

Answer:

maybe use helpful, caring, kind, encouraging

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What is the third quartile of this data set?<br> 21, 24, 25, 28, 29, 35, 37, 43, 44
Ronch [10]

Answer:

43.5

Step-by-step explanation:

6 0
3 years ago
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Sharon wants to buy a shirt that costs $50. the sales tax is 5%. how much is the sales tax? what is her total cost for the shirt
maria [59]
5% of $50 = 5/100 x 50 = $2.50
Total = 50 + 2.50 = $52.50

The saes tax is $2.50; her total cost of the shirt is $52.50
6 0
3 years ago
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Why must one leg of a triangle that the Pythagorean triple represents need an even-numbered length?
pashok25 [27]

Most of the square numbers are even.so if we need to find root over we need an even number to convert it to perfect square.

See the example below

  • P=3
  • B=4
  • H=5

\\ \ast\bull\rm\longmapsto H^2=P^2+B^2

\\ \ast\bull\rm\longmapsto H^2=3^2+4^2

\\ \ast\bull\rm\longmapsto H^2=9+16

\\ \ast\bull\rm\longmapsto H^2=25

\\ \ast\bull\rm\longmapsto H=\sqrt{25}

\\ \ast\bull\rm\longmapsto H=5

Some other triplets are

  • (6,8,10)
  • (5,12,13)
  • (8,15,17)
6 0
2 years ago
Find the value of x for following problems
leva [86]

Answer:

find the x in my bag ;D

5 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
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