Solve by completing the square. <br> x2+12x+4=0

Oil is pumped continuously from a well at a rate proportional to the amount of oil left in the well. Initially there were millio

JulijaS [17]

Answer:

The amount of oil was decreasing at 69300 barrels, yearly

Step-by-step explanation:

Given

$Initial =1\ million$

$6\ years\ later = 500,000$

Required

At what rate did oil decrease when 600000 barrels remain

To do this, we make use of the following notations

t = Time

A = Amount left in the well

So:

$\frac{dA}{dt} = kA$

Where k represents the constant of proportionality

$\frac{dA}{dt} = kA$

Multiply both sides by dt/A

$\frac{dA}{dt} * \frac{dt}{A} = kA * \frac{dt}{A}$

$\frac{dA}{A} = k\ dt$

Integrate both sides

$\int\ {\frac{dA}{A} = \int\ {k\ dt}$

$ln\ A = kt + lnC$

Make A, the subject

$A = Ce^{kt}$

$t = 0\ when\ A =1\ million$ i.e. At initial

So, we have:

$A = Ce^{kt}$

$1000000 = Ce^{k*0}$

$1000000 = Ce^{0}$

$1000000 = C*1$

$1000000 = C$

$C =1000000$

Substitute $C =1000000$ in $A = Ce^{kt}$

$A = 1000000e^{kt}$

To solve for k;

$6\ years\ later = 500,000$

i.e.

$t = 6\ A = 500000$

So:

$500000= 1000000e^{k*6}$

Divide both sides by 1000000

$0.5= e^{k*6}$

Take natural logarithm (ln) of both sides

$ln(0.5) = ln(e^{k*6})$

$ln(0.5) = k*6$

Solve for k

$k = \frac{ln(0.5)}{6}$

$k = \frac{-0.693}{6}$

$k = -0.1155$

Recall that:

$\frac{dA}{dt} = kA$

Where

$\frac{dA}{dt}$ = Rate

So, when

$A = 600000$

The rate is:

$\frac{dA}{dt} = -0.1155 * 600000$

$\frac{dA}{dt} = -69300$

<em>Hence, the amount of oil was decreasing at 69300 barrels, yearly</em>

7 0

3 years ago

What is the answer to this calculus problem???<br><img src="https://tex.z-dn.net/?f=%20%5Cfrac%7Bd%7D%7Bdx%7D%20%28e%20%7B%7D%5E

Len [333]

Answer:

see explanation

Step-by-step explanation:

Differentiate using the product rule

Given y = f(x)g(x), then

$\frac{dy}{dx}$ = f(x). g'(x) + g(x). f'(x)

here f(x) = $e^{x}$ ⇒ f'(x) = $e^{x}$

g(x) = cosx ⇒ g'(x) = - sinx

Hence

$\frac{dy}{dx}$ = $e^{x}$ (- sinx) + cosx $e^{x}$

= $e^{x}$ cosx - $e^{x}$ sinx

3 0

3 years ago

Which of the following, to the nearest tenth, is a solution of f(x) = g (x) if f (x) = in (x + 5 ) - 1 and g(x) =x^3 - 2x+1

Dovator [93]

Answer:

c

Step-by-step explanation:

5 0

3 years ago

Please someone help, give the right answer it’s important

dangina [55]

9514 1404 393

Answer:

(4)

Step-by-step explanation:

The fact that GH is the perpendicular bisector of EF means EM = FM and all of the angles at M are congruent.

It does not say anything about the relative sizes of HM and GM, so we cannot conclude HM and GM are congruent.

5 0

3 years ago

Consider the function f(x) = x(x-4.

jeyben [28]

Step-by-step explanation:

point (c-2, y) lies on the graph of f(x)=x(x-4)f(x)=x(x−4) .

Step-by-step explanation:

Given function f(x)=x(x-4)f(x)=x(x−4) also point (2+ c,y) is on the graph of f(x) ,

We have to find out of given point which point will also be on the graph of f(x).

Consider the given function f(x)=x(x-4)f(x)=x(x−4)

f(x)=x(x-4)f(x)=x(x−4) can be rewritten f(x)=x^2-4xf(x)=x

2

−4x

Now we substitute the given point (2+ c, y) in the function given ,

we have,

f(x)=y=x(x-4)f(x)=y=x(x−4)

put for x as 2+c , we have,

\Rightarrow y=(2+c)(2+c-4)⇒y=(2+c)(2+c−4)

Solve, we get

\Rightarrow y=(2+c)(c-2)⇒y=(2+c)(c−2)

Thus, both point (2+c, y) and (c-2, y) lies on the graph of f(x)=x(x-4)f(x)=x(x−4)

Thus, option (1) is correct.

3 0

3 years ago

Solve by completing the square. ​ x2+12x+4=0

Solve by completing the square. x2+12x+4=0