Convert 16/100 to decimal.

Un viajero va a la ciudad cada 18 días y otro cada 24 días. Hoy han coincidido los dos ahi . ¿Dentro de cuántos días volverán a

spayn [35]

Si tengo la razon seria durante 84 dias.

8 0

3 years ago

Please help me, I don't know how to do this...

Lunna [17]

Answer:

A. 5

Step-by-step explanation:

The triangle will be a right triangle when the side lengths satisfy the Pythagorean theorem: the sum of the squares of the legs is equal to the square of the hypotenuse.

(x +1)^2 +x^2 = (√61)^2

x^2 +2x +1 +x^2 = 61 . . . eliminate parentheses

2x^2 +2x = 60 . . . . . . .subtract 1, collect terms

x^2 +x -30 = 0 . . . . . divide by 2, subtract 30

(x +6)(x -5) = 0 . . . . factor the quadratic

x = -6 or +5

The solution is x = 5.

_____

Side lengths cannot be negative. Solution values are values of x that make the factors zero. x+6=0 when x=-6, for example.

8 0

2 years ago

Find the circumference of a circle with a diameter of 9 in.

Arturiano [62]

Answer:

28.27 inches

Step-by-step explanation:

circumference of a circle = πd

9π = 28.27

8 0

2 years ago

A 75-gallon tank is filled with brine (water nearly saturated with salt; used as a preservative) holding 11 pounds of salt in so

Debora [2.8K]

Let $A(t)$ = amount of salt (in pounds) in the tank at time $t$ (in minutes). Then $A(0) = 11$ .

Salt flows in at a rate

$\left(0.6\dfrac{\rm lb}{\rm gal}\right) \left(3\dfrac{\rm gal}{\rm min}\right) = \dfrac95 \dfrac{\rm lb}{\rm min}$

and flows out at a rate

$\left(\dfrac{A(t)\,\rm lb}{75\,\rm gal + \left(3\frac{\rm gal}{\rm min} - 3.25\frac{\rm gal}{\rm min}\right)t}\right) \left(3.25\dfrac{\rm gal}{\rm min}\right) = \dfrac{13A(t)}{300-t} \dfrac{\rm lb}{\rm min}$

where 4 quarts = 1 gallon so 13 quarts = 3.25 gallon.

Then the net rate of salt flow is given by the differential equation

$\dfrac{dA}{dt} = \dfrac95 - \dfrac{13A}{300-t}$

which I'll solve with the integrating factor method.

$\dfrac{dA}{dt} + \dfrac{13}{300-t} A = \dfrac95$

$-\dfrac1{(300-t)^{13}} \dfrac{dA}{dt} - \dfrac{13}{(300-t)^{14}} A = -\dfrac9{5(300-t)^{13}}$

$\dfrac d{dt} \left(-\dfrac1{(300-t)^{13}} A\right) = -\dfrac9{5(300-t)^{13}}$

Integrate both sides. By the fundamental theorem of calculus,

$\displaystyle -\dfrac1{(300-t)^{13}} A = -\dfrac1{(300-t)^{13}} A\bigg|_{t=0} - \frac95 \int_0^t \frac{du}{(300-u)^{13}}$

$\displaystyle -\dfrac1{(300-t)^{13}} A = -\dfrac{11}{300^{13}} - \frac95 \times \dfrac1{12} \left(\frac1{(300-t)^{12}} - \frac1{300^{12}}\right)$

$\displaystyle -\dfrac1{(300-t)^{13}} A = \dfrac{34}{300^{13}} - \frac3{20}\frac1{(300-t)^{12}}$

$\displaystyle A = \frac3{20} (300-t) - \dfrac{34}{300^{13}}(300-t)^{13}$

$\displaystyle A = 45 \left(1 - \frac t{300}\right) - 34 \left(1 - \frac t{300}\right)^{13}$

After 1 hour = 60 minutes, the tank will contain

$A(60) = 45 \left(1 - \dfrac {60}{300}\right) - 34 \left(1 - \dfrac {60}{300}\right)^{13} = 45\left(\dfrac45\right) - 34 \left(\dfrac45\right)^{13} \approx 34.131$

pounds of salt.

7 0

1 year ago

Find the area of a trapezoid with b1 = 3 cm, b2 = 22 cm, and h = 6 cm.

Molodets [167]

Area of trapezoid = Average length of parallel lines x height

3+22=25
25/2= 12.5
12.5x6=75cm^2

6 0

2 years ago

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