Solve the system by elimination

27a+1/2b=4 solve for a

sladkih [1.3K]

Answer:

a = 4/27 - b/54

Step-by-step explanation:

27a+1/2b=4

Subtract 1/2 b from each side

27a+1/2b - 1/2 b=4-1/2 b

27a = 4 - 1/2 b

Divide each side by 27

27a/27 = 4/27 - 1/2 b/27

a = 4/27 - b/54

5 0

3 years ago

Read 2 more answers

Solve 2w^2– 16w = 12W – 48 by factoring. Select the solution(s)

UkoKoshka [18]

Answer:

w=12 or w=2

Step-by-step explanation:

2w^2-16w=12w-48

2w^2-28w=-48

2w^2-28w+48=0

2(w^2-14w+24)=0

2(w-12)(w-2)=0

(w-12)(w-2)=0

w=12 or w=2

3 0

3 years ago

The rent for an office space is $674.30 per month. The office is 306 1/2 square feet in area.What is the monthly cost per square

Rufina [12.5K]

The answer is B because 674.30 divided by 306.5 equals 2.2 or $2.20

3 0

3 years ago

Natsumi went on a business lunch. The pretax bill was $55.60. She decided

igomit [66]

8.25% of 55.60 = 4.58700

We round that number to 4.59

(I am assuming the tip is based on the amount of the pretax bill.)

15% of 55.60 = 8.34

So $55.60 + $4.59 + 8.34 = $68.53

The bill would be $68.53.

5 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