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vlada-n [284]
3 years ago
10

What is the first step to isolate the variable term on one side of the equation?

Mathematics
1 answer:
Aleksandr-060686 [28]3 years ago
4 0

Answer:

The basic technique to isolate a variable is to “do something to both sides” of the equation, such as add, subtract, multiply, or divide both sides of the equation by the same number. By repeating this process, we can get the variable isolated on one side of the equation.

Step-by-step explanation:

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Please help me with this
elena-s [515]

Answer:

9 p.m.

Step-by-step explanation:

Lowest common multiple (LCM) of 6 and 4.

LCM of 6, 4 = 2 × 2 × 3 = 12

After 12 hours 9 A.M. = 9 P.M.

After 12 hours at 9 P.M. she will take both medicines together again.

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2 and 3/8 minus 3/4 is greater than or less than 2?
ohaa [14]
It is less than 2. I hope this helps!
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Explain how you can round 25.691 to the greatest place.
nadezda [96]
You have to round it to the nearest hundreds
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3 years ago
Where does a hole occur in the graph of y = x(x+2) / x(x-3)
Morgarella [4.7K]

Answer:

A

Step-by-step explanation:

Given

y = \frac{x(x+2)}{x(x-3)} ← cancel x on numerator/ denominator

 = \frac{x+2}{x-3}

Cancelling the factor x, leaves a hole in the graph at x = 0

7 0
3 years ago
The rate of change of the number of mountain lions N(t) in a population is directly proportional to 725 - N(t), where t is the t
wolverine [178]

Answer:

a) The implied differential equation is \frac{dN}{725 - N} = kdt

b) The general equation is N = 725 - C e^{-kt}

c) The particular equation is N = 725 - 325 e^{-0.49t}

d) The population when t = 5, N(5) = 697 = 700( to the nearest 50)

Step-by-step explanation:

The rate of change of N(t) can be written as dN/dt

According to the question, \frac{dN}{dt} \alpha (725 - N(t))

\frac{dN}{dt} = k (725 - N)\\\frac{dN}{725 - N} = kdt

Integrating both sides of the equation

\int {\frac{1 }{725 - N}} \, dN  = \int {k} \, dt\\- ln (725 - N) = kt + C\\ ln (725 - N) = -(kt + C)\\725 - N = e^{-(kt + C)} \\725 - N = e^{-kt} * e^{-C} \\725 - N = C e^{-kt}\\N = 725 - C e^{-kt}

When t = 0, N = 400

400 = 725 - C e^{-k*0}\\400 = 725 - C\\C = 725 - 400\\C = 325

When t = 3,  N = 650

650 = 725 - (325 * e^{-3k})\\325 * e^{-3k} = 75\\e^{-3k} = 75/325\\e^{-3k} = 0.23\\-3k = ln 0.23\\-3k = -1.47\\k = 1.47/3\\k = 0.49

The equation for the population becomes:

N = 725 - 325 e^{-0.49t}

At t = 5, the population becomes:

N = 725 - 325 e^{-0.49*5}\\N = 725 - 325 e^{-2.45}\\N = 696.95\\N(5) = 697

N(5) = 700 ( to the nearest 50)

6 0
3 years ago
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