If 3x−y=12, what is the value of <br>8x<br>--<br>2y

Make a friend right now 10 points for whoever douse

Afina-wow [57]

Hand me dem points rn

5 0

3 years ago

A baker uses 2.327 kilograms of blueberries for muffins. There are four packages at the fruit market. Which amount is closest to

irinina [24]

Answer:

2.33 kg

Step-by-step explanation:

If the baker needs to use 2.327 kilograms of blueberries, we are going to subtract the amount of the packages to the amount he needs and we'll see which is the smallest number.

In the case of 2.3, 2.327 - 2.3 = .027

In the case of 2.42, 2.327 - 2.42 = -0.09

In the case of 2.33, 2.327 - 2.33 = -0.003

In the case of 2.4, 2.327 - 2.4 = -0.07

Therefore, the closest amount to 2.327 is the 2.33 package.

4 0

3 years ago

Some good f the Aliens have hairs on their heads that change color according to their emotions. If they're pleased hairs turn or

jeka94

7 because you divide the 27 from 3.

3 0

3 years ago

Consider the quadratic function f(x) = x2 – 5x + 12. Which statements are true about the function and its graph? Select three op

klasskru [66]

The graph of the quadratic function f(x)=x² - 5x + 12 is a parabola which opens up.

<h3>What is an equation?</h3>

An equation is an expression that shows the relationship between two or more numbers and variables.

The standard form of a quadratic equation is:

y = ax² + bx + c

The graph of a quadratic equation is a parabola.

The graph of the quadratic function f(x)=x² - 5x + 12 is a parabola which opens up.

Find out more on equation at: brainly.com/question/2972832

#SPJ1

4 0

2 years ago

Litter such as leaves falls to the forest floor, where the action of insects and bacteria initiates the decay process. Let A be

Travka [436]

Answer:

D = L/k

Step-by-step explanation:

Since A represents the amount of litter present in grams per square meter as a function of time in years, the net rate of litter present is

dA/dt = in flow - out flow

Since litter falls at a constant rate of L grams per square meter per year, in flow = L

Since litter decays at a constant proportional rate of k per year, the total amount of litter decay per square meter per year is A × k = Ak = out flow

So,

dA/dt = in flow - out flow

dA/dt = L - Ak

Separating the variables, we have

dA/(L - Ak) = dt

Integrating, we have

∫-kdA/-k(L - Ak) = ∫dt

1/k∫-kdA/(L - Ak) = ∫dt

1/k㏑(L - Ak) = t + C

㏑(L - Ak) = kt + kC

㏑(L - Ak) = kt + C' (C' = kC)

taking exponents of both sides, we have

$L - Ak = e^{kt + C'} \\L - Ak = e^{kt}e^{C'}\\L - Ak = C"e^{kt} (C" = e^{C'} )\\Ak = L - C"e^{kt}\\A = \frac{L}{k} - \frac{C"}{k} e^{kt}$

When t = 0, A(0) = 0 (since the forest floor is initially clear)

$A = \frac{L}{k} - \frac{C"}{k} e^{kt}\\0 = \frac{L}{k} - \frac{C"}{k} e^{k0}\\0 = \frac{L}{k} - \frac{C"}{k} e^{0}\\\frac{L}{k} = \frac{C"}{k} \\C" = L$

$A = \frac{L}{k} - \frac{L}{k} e^{kt}$

So, D = R - A =

$D = \frac{L}{k} - \frac{L}{k} - \frac{L}{k} e^{kt}\\D = \frac{L}{k} e^{kt}$

when t = 0(at initial time), the initial value of D =

$D = \frac{L}{k} e^{kt}\\D = \frac{L}{k} e^{k0}\\D = \frac{L}{k} e^{0}\\D = \frac{L}{k}$

4 0

3 years ago

If 3x−y=12, what is the value of 8x--2y