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Aneli [31]
3 years ago
11

Given then functions f(x)= 2x -1 and g(x) =2x +4, which operation results in the largest coefficient on the x term?

Mathematics
2 answers:
denpristay [2]3 years ago
8 0

Answer:multiplication.


Step-by-step explanation:



RideAnS [48]3 years ago
5 0
When two equations are given, there are many operations in which you can do involving both as long as the variables (independent and dependent) are similar to both equations. In this case, we can directly apply operations to both equations since both functions are functions of x. The largest coefficient of the x term would be achieved by multiplication since using distribution property, the coefficient of x would be 6. Subtraction or addition would only lead to less number (0 and 4 respectively). Division is not applicable as there are no factors similar to both functions which can be cancelled to each other. Answer then to this problem is multiplication.
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A pond forms as water collects in a conical depression of radius a and depth h. Suppose that water flows in at a constant rate k
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Answer:

a. dV/dt = K - ∝π(3a/πh)^⅔V^⅔

b. V = (hk^3/2)/[(∝^3/2.π^½.(3a))]

The small deviations from the equilibrium gives approximately the same solution, so the equilibrium is stable.

c. πa² ≥ k/∝

Step-by-step explanation:

a.

The rate of volume of water in the pond is calculated by

The rate of water entering - The rate of water leaving the pond.

Given

k = Rate of Water flows in

The surface of the pond and that's where evaporation occurs.

The area of a circle is πr² with ∝ as the coefficient of evaporation.

Rate of volume of water in pond with time = k - ∝πr²

dV/dt = k - ∝πr² ----- equation 1

The volume of the conical pond is calculated by πr²L/3

Where L = height of the cone

L = hr/a where h is the height of water in the pond

So, V = πr²(hr/a)/3

V = πr³h/3a ------ Make r the subject of formula

3aV = πr³h

r³ = 3aV/πh

r = ∛(3aV/πh)

Substitute ∛(3aV/πh) for r in equation 1

dV/dt = k - ∝π(∛(3aV/πh))²

dV/dt = k - ∝π((3aV/πh)^⅓)²

dV/dt = K - ∝π(3aV/πh)^⅔

dV/dt = K - ∝π(3a/πh)^⅔V^⅔

b. Equilibrium depth of water

The equilibrium depth of water is when the differential equation is 0

i.e. dV/dt = K - ∝π(3a/πh)^⅔V^⅔ = 0

k - ∝π(3a/πh)^⅔V^⅔ = 0

∝π(3a/πh)^⅔V^⅔ = k ------ make V the subject of formula

V^⅔ = k/∝π(3a/πh)^⅔ -------- find the 3/2th root of both sides

V^(⅔ * 3/2) = k^3/2 / [∝π(3a/πh)^⅔]^3/2

V = (k^3/2)/[(∝π.π^-⅔(3a/h)^⅔)]^3/2

V = (k^3/2)/[(∝π^⅓(3a/h)^⅔)]^3/2

V = (k^3/2)/[(∝^3/2.π^½.(3a/h))]

V = (hk^3/2)/[(∝^3/2.π^½.(3a))]

The small deviations from the equilibrium gives approximately the same solution, so the equilibrium is stable.

c. Condition that must be satisfied

If we continue adding water to the pond after the rate of water flow becomes 0, the pond will overflow.

i.e. dV/dt = k - ∝πr² but r = a and the rate is now ≤ 0.

So, we have

k - ∝πa² ≤ 0 ---- subtract k from both w

- ∝πa² ≤ -k divide both sides by - ∝

πa² ≥ k/∝

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