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3 years ago

Find the 11th term for : 64, -32, 16, -8A . -.625B. 0.625C. .03125D. -.03125

Mathematics

1 answer:

MA_775_DIABLO [31]3 years ago

8 0

Answer:

0.0625 is what i got

Step-by-step explanation:

64, -32, 16, --8, 4, -2, 1, -.5, .25, -.125, 0.0625

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What are the 2 square roots of 0.0625

Fofino [41]

0.03125 and 0.015625

hope this helps but im probably wrong

7 0

2 years ago

selena walks home to school each mm morning and back home each afternoon. altogether she walks 2/3 mile wach day. how far does s

LenaWriter [7]

Answer:

One days trip of to school from home and back home from school is 2/3 of a mile. We want to know how far it is to school from her house.

To solve this, we simply need to take half of the total distance (2/3)

\frac{2}{3} / 2

Next, we need to turn the 2 into a fraction. Every whole number can be made into a fraction by putting it over 1.

\frac{2}{3} / \frac{2}{1}

Because we are dividing, we need to invert the second fraction and then multiply.

\frac{2}{3} * \frac {1}{2}

Next, we multiply the top of the first fraction by the top of the second and the bottom of the first fraction by the bottom of the second.

\frac{2}{6}

Once you reduce, you get:

\frac{1}{3}

4 0

3 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

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