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3 years ago

O curso de inglês de Carolina teve duração de 4 semestres e 2 meses . quantos tempo durou esse curso de Carolina ?

Mathematics

1 answer:

Afina-wow [57]3 years ago

5 0

Answer:

a 26 meses gracias por los puntos

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Hello can anyone help me ? here the question

lesantik [10]

Answer:

A. 1/5k - 2/3j and -2/3j +1/5k

Step-by-step explanation:

A. 1/5k - 2/3j and -2/3j +1/5k

B. 1/5k - 2/3j and -1/5k +2/3j

There is a change in the signs of each term

1/5k changed to -1/5k

-2/3j changed to +2/3j

Not equivalent

C. 1/5k - 2/3j and 1/5j - 2/3k

There is a change in the variables

1/5k changed to 1/5j

-2/3j changed to -2/3k

D. 1/5k - 2/3j and 2/3j - 1/5k

The is a change in the signs of each term

1/5k changed to -1/5k

-2/3j changed to +2/3j

The only equivalent expression is

A. 1/5k - 2/3j and -2/3j +1/5k

3 0

3 years ago

Consider the linear equation y=4x-7 find the value of y that makes the given ordered pair a solution to the equation (3, y) Y=

olchik [2.2K]

Y = 4x - 7 is a line with a slope of 4 and y-intercept of -7.

plug in any X value, for example 1 and you get the following

y = 4(1) - 7

y = -3

(1,-3) is a point that satisfies the equation y = 4x - 7

3 0

3 years ago

How do I find the values of x and y?

zepelin [54]

All 3 angles =180...

3 0

4 years ago

What are the zeros of f(x)=x^2-10+25

BartSMP [9]

Answer:
-5

Explanation:
Because it’s right

7 0

3 years ago

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

2 years ago