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sesenic [268]
3 years ago
14

SOMEONE JUST PLEASE ANSWER THIS ASAP FOR BRAINLIEST!!!

Mathematics
2 answers:
vekshin13 years ago
7 0
2k^2×(k+7)×(k-5)
that ur answer
Kazeer [188]3 years ago
6 0

2k^{2}(k-5)(k-7)

Hope I could help!

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Which of the following is the best estimate of the square root of 90
guapka [62]

Answer:

9.5

Step-by-step explanation:

9.5x9.5= 90.25 which is the closest you can get without going into further decimals

5 0
3 years ago
Solve x2 = 12x - 15 by completing the square. Which is the solution set of the equation?
Novosadov [1.4K]

Answer:

(6-√21,  6+√21)

Step-by-step explanation:

x^2 - 12x = -15

(x - 6)^2 - 36 = -15

(x - 6)^2 = 21

x - 6 = ± √21

x = 6  ± √21

8 0
3 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
I will give you BRAINLIEST for the correct answer
Nadya [2.5K]

Answer:

a or b

Step-by-step explanation:

if I did the math right it's a but I tried another way and got b, so I'm sorry if that was no help at all

6 0
3 years ago
Read 2 more answers
Ben played at a friends house for 2 hours and 35 minutes.Later he played at a park.He played for a total of 3 hours 52 minutes t
rjkz [21]
1hour and 17 minutes
8 0
3 years ago
Read 2 more answers
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