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1 year ago

What is the discriminant of the quadratic equation 3 – 4x = –6x2? –68 –56 76 88

Mathematics

1 answer:

Ulleksa [173]1 year ago

8 0

Answer:

- 56

Step-by-step explanation:

Given a quadratic equation in standard form

ax² + bx + c = 0 ( a ≠ 0 )

Then the discriminant Δ = b² - 4ac

Given

3 - 4x = - 6x² ( add 6x² to both sides )

6x² - 4x + 3 = 0 ← in standard form

with a = 6, b = - 4 and c = 3, thus

b² - 4ac

= (- 4)² - (4 × 6 × 3) = 16 - 72 = - 56

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After years of maintaining a steady population of 32,000, the population of a town begins to grow exponentially. After 1 year an

galben [10]

Answer:

$y=32000(1+0.08)^x$

Step-by-step explanation:

Exponential growth function is $y=a(1+r)^x$

Where 'a' is the initial population

r is the rate of growth and x is the time period in years

a steady population of 32,000. So initial population is 32,000

an increase of 8% per year. the rate of increase is 8% that is 0.08

a= 32000 and r= 0.08

Plug in all the values in the general equation

$y=a(1+r)^x$

$y=32000(1+0.08)^x$

$y=32000(1+0.08)^x$

4 0

2 years ago

Read 2 more answers

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/∝

5 0

2 years ago

Point C' (-4,-3) is the image of C (-2,-3) under a translation

inysia [295]

That might be a dilation but im not entirely sure

7 0

2 years ago

Match the following guess solutions ypyp for the method of undetermined coefficients with the second-order nonhomogeneous linear

Sunny_sXe [5.5K]

Answer:

Step-by-step explanation:

1 ) Given that

$(d^2y/dx^2) + 4y = x - x^2 + 20\\\\ (d^2y/dx^2) + 4y = - x^2 + x + 20$

For a non homogeneous part $- x^2 + x + 20$ , we assume the particular solution is

$y_p(x) = Ax^2 + Bx + C$

2 ) Given that

$d^2y/dx^2 + 6dy/dx + 8y = e^{2x}$

For a non homogeneous part $e^{2x}$ , we assume the particular solution is

$y_p(x) = Ae^{2x}$

3 ) Given that

y′′ + 4y′ + 20y = −3sin(2x)

For a non homogeneous part −3sin(2x) , we assume the particular solution is

$y_p(x) = Acos(2x)+Bsin(2x)$

4 ) Given that

y′′ − 2y′ − 15y = 3xcos(2x)

For a non homogeneous part 3xcos(2x) , we assume the particular solution is

$y_p(x) = (Ax+B)cos2x+(Cx+D)sin2x$

4 0

1 year ago

An ice cream store has two new flavors: Fantasy and Ecstasy. Each barrel of Fantasy requires 4

Y_Kistochka [10]

Answer:

Fantasy: 3 barrels
Ecstasy: 1 barrel

Step-by-step explanation:

Given

Fantasy uses 4 lb of nuts, 3 lb of chocolate, for a profit of $50

Ecstasy uses 4 lb of nuts, 1 lb of chocolate, for a profit of $40

In stock are 16 lb of nuts, 10 lb of chocolate

Find

amount of each to maximize profit

Solution

Let x and y represent barrels of Fantasy and Ecstasy, respectively. Then the limitations on production are ...

4x +4y ≤ 16 . . . lb of nuts

3x +y ≤ 10 . . . . lb of chocolate

We want to maximize

50x +40y

The graph shows the feasible region. Its vertices are ...

(0, 4), (3, 1), (3.33, 0)

Profit is maximized at $190 when production is 3 barrels of Fantasy and 1 barrel of Ecstasy.

8 0

2 years ago