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ivann1987 [24]
2 years ago
5

-8z+12=-12 plz help asap what is the answer and what does z stand for??!??

Mathematics
2 answers:
Sveta_85 [38]2 years ago
6 0

Answer:

z is a variable so it can be any number.

z=-3

if you do -8 times -3 that equals 24.

the answer is -3

Step-by-step explanation:

vichka [17]2 years ago
4 0

Answer:

<em>z </em>stands for 3.

Step-by-step explanation:

First, subtract 12 from both sides.

-8z + 12 - 12 = -12 - 12

-8z = -24

Next, let's put brackets [] between both numbers. Any number inside of these becomes positive.

[-8z] = [-24]

8z = 24

Finally, let's divide doth sides by 8!

8z/8 = 24/8

z = 3

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After the data was recorded, Leah realized that she left out a student who handed out 42 flyers. How would the mean number of fl
Yuliya22 [10]

Answer:

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Step-by-step explanation:

8 0
3 years ago
Find the product.<br><br> (-5a ^2)^3·a ^5<br><br> A)-15a10<br> B)-125a11<br> C)15a6<br> D)-125a10
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  B)  -125a^11

Step-by-step explanation:

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5 0
2 years ago
Read 2 more answers
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Answer:

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Step-by-step explanation:

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3 0
3 years ago
Read 2 more answers
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

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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

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

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

4 0
3 years ago
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