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2 years ago

Solve the system using elimination. x + 2y = –6 3x + 8y = –20 (–4, –1) (–4, 4) (–1, –4) (3, 1)

Mathematics

2 answers:

Orlov [11]2 years ago

6 0

$x + 2y = - 6 > > > 3x = - 18 - 6y$
$- 18 - 6y + 8y = - 20 > > > 2y = - 2 > > y = - 1$
so x=4

LiRa [457]2 years ago

3 0

Answer: The required solution is x = -4 and y = -1.

Step-by-step explanation: We are given to solve the following system of equations by the method of Elimination :

$x+2y=-6~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)\\\\3x+8y=-20~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(ii)$

Multiplying equation (i) by 3 , we have

$3(x+2y)=-6\times3\\\\\Rightarrow 3x+6y=-18~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(iii)$

Subtracting equation (iii) from equation (ii), we get

$(3x+8y)-(3x+6y)=-20-(-18)\\\\\Rightarrow 2y=-20+18\\\\\Rightarrow 2y=-2\\\\\Rightarrow y=-\dfrac{2}{2}\\\\\Rightarrow y=-1.$

From equation (i), we get

$x+2\times(-1)=-6\\\\\Rightarrow x-2=-6\\\\\Rightarrow x=-6+2\\\\\Rightarrow x=-4.$

Thus, the required solution is x = -4 and y = -1.

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1 year ago

Exhibit 11-2 We are interested in determining whether the variances of the sales at two music stores (A and B) are equal. A samp

Fiesta28 [93]

Answer:

Option A , should not be rejected

Step-by-step explanation:

From the question we are told that:

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1 year ago

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

Scrat [10]

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:

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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