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julia-pushkina [17]
2 years ago
13

Please Help Will mark brainliest

Mathematics
1 answer:
Rudik [331]2 years ago
7 0

Answer:

○ \displaystyle -6x^2 + 6x + 5

Step-by-step explanation:

Since the divisor is in the form of \displaystyle x - c,use what is called Synthetic Division. Remember, in this formula, -c gives you the OPPOCITE TERMS OF WHAT THEY <em>REALLY</em> ARE, so do not forget it. Anyway, here is how it is done:

2| −6 18 −7 −10

↓ −12 12 10

________________

−6 6 5 0 \displaystyle \hookrightarrow -6x^2 + 6x + 5

You start by placing the <em>c</em> in the top left corner, then list all the coefficients of your dividend [−6x³ + 18x² - 7x - 10]. You bring down the original term closest to <em>c</em> then begin your multiplication. Now depending on what symbol your result is tells you whether the next step is to subtract or add, then you continue this process starting with multiplication all the way up until you reach the end. Now, when the last term is 0, that means you have <em>no remainder</em>. Finally, your quotient is <em>one degree less</em> than your dividend, so that −6 in your quotient can be a \displaystyle -6x^2,the \displaystyle 6xfollows right behind it, then you end it off with \displaystyle 5,giving you the quotient of \displaystyle -6x^2 + 6x + 5.

I am joyous to assist you at any time.

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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
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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}.

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Answer:

the "?" represents the number 175

Step-by-step explanation:

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so, you would multiply the 25 by 7 to get 175

hope this helps!

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