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Leya [2.2K]
3 years ago
10

-6x+3y=-39 -6x-3y=-21 solve the system of equations

Mathematics
2 answers:
zimovet [89]3 years ago
5 0
The answer is the one on top
solmaris [256]3 years ago
4 0

Answer:

x=5,y=-3

Step-by-step explanation:

\left \{ {{-6x+3y=-39} \atop {-6x-3y=-21}} \right. \\-12x=-60\\12x=60\\x=\frac{60}{12} \\x=5

-6(5)+3y=-39\\-30+3y=-39\\3y=-39+30\\3y=-9\\y=-\frac{9}{3} \\y=-3

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The differential equation in Example 3 of Section 2.1 is a well-known population model. Suppose the DE is changed to dP dt = P(a
LuckyWell [14K]

Answer:

Decreases

Step-by-step explanation:

We need to determine the integral of the DE;

dP/dt=P(aP-b)

dP=P(aP-b)dt

1/(dP^2-bP)dP=dt

We can solve this by integration by parts on the left side. We expand the fraction 1/P²:

1/(d-b/P)\cdot{P^2} dP

let

u=d-b/P

du/dP=b/P^2

dP=\int\limits {P^2/b} \, du

P=lnu/b

Substitute u in:

P=ln(d-b/P)/b

Therefore the equation is:

ln(d-b/P)/b=t

We simplify:

d-b/P=e^b^t

P=b/(d-e^b^t)

As t increases to infinity P will decrease

6 0
3 years ago
Which equation is in point-slope form for the given point and slope?
faust18 [17]

Answer: y-7 = 4(x+3), choice B

============================================

Point slope form is generally

y-y1 = m(x-x1)

we have m as the slope and (x1,y1) as the point this line goes through. In this case,

m = 4

(x1,y1) = (-3,7) so x1 = -3 and y1 = 7

So,

y-y1 = m(x-x1)

y - 7 = 4(x - (-3))

y - 7 = 4(x + 3)

which is why choice B is the answer

6 0
3 years ago
Read 2 more answers
Paige has to make invitation cards for a party. If she can make 18 cards in 2 days, how many cards can she make in 30 days?
mylen [45]
18/2=x/30
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5 0
3 years ago
The radius of a circle is changing at the rate of 1/π inches per second. At what rate, in square inches per second, is the circl
MAXImum [283]

Answer:

10 square inches per second.

Step-by-step explanation:

The radius of the circle is given by the equation:

r(t) = (1/π  in/s)*t

Where time in seconds.

Remember that the area of a circle of radius R is written as:

A = π*R^2

Then the area of our circle will be:

A(t) = π*( (1/π  in/s)*t)^2 = π*(1/π  in/s)^2*(t)^2

Now we want to find the rate of change (the first derivation of the area) when the radius is equal to 5 inches.

Then the first thing we need to do is find the value of t such that the radius is equal to 5 inches.

r(t) = 5 in =  (1/  in/s)*t

       5in*(π s/in) = t

        5*π s = t

So the radius will be equal to 5 inches after 5*π seconds, let's remember that.

Now let's find the first derivate of A(t)

dA(t)/dt = A'(t) = 2*(π*(1/π  in/s)^2*t = (2*π*t)*(1/π  in/s)^2

Now we need to evaluate this in the time such that the radius is equal to 5 inches, we will get:

A'(5*π s) = (2*π*5*π s)*((1/π  in/s)^2

              = (10*π^2  s)*(1/π^2  in^2/s^2) = 10 in^2/s

The rate of change is 10 square inches per second.

4 0
3 years ago
PLEASE HELP 11 POINTS
Free_Kalibri [48]

Answer:

It should be 87.92

Step-by-step explanation:

4 0
3 years ago
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