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Murrr4er [49]
3 years ago
11

Solve for x. x/4 + 6 = 10

Mathematics
2 answers:
Luden [163]3 years ago
7 0
X=4 because 6-10=4 and it makes sense
Papessa [141]3 years ago
5 0

Answer:

16

Step-by-step explanation:

<em>16/4=4 so replace x for 4</em>

<em>Next you solve 16/4+6=10</em>

<em>Simplify 4+6=10</em>

<em>Make sure the question is correct. 10-6=4</em>

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Adding a negative is the same as subtracting a positive

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   -9.2           +              -6.7

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The Sullivan household wants build a patio deck in the shape of a 45-45-90 triangle in a nice corner section of their backyard .
dedylja [7]

Answer:

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Step-by-step explanation:

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4 0
4 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
Help thank you!!!!!!!
Vesnalui [34]

v = \sqrt{4900} + \sqrt{8100} = 70 + 90 = 160

Answer: D. 160

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