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4 years ago

How to do lond division 0.6÷0.0024

Mathematics

1 answer:

Kryger [21]4 years ago

3 0

So...notice first off, the 0.6 has 1 decimal, the 0.0024 has 4 decimals

what happens is, the dot goes poof BUT, you need to add as many zeros to each number as the other's decimals.

the 0.6 has one decimal, so 0.0024 gets an added zero and loses the dot, to become 000240.

the 0.0024 has four decimals, thus the 0.6 gets added 4 zeros then, and loses the dot, to become 060000

and then you divided that, those are the dividend and divisor.

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Convert to scientific notation.<br> 16,120

eduard

Answer:

$1.612\times10^4$

Step-by-step explanation:

In order to find the notation of a number find the total number of 0's that the number has, and realize that all whole numbers have a invisible decimal point.

$1.612\times10^4$

Exponent is positive to move the decimal point to the right four times

$1.6.1.2.0.$

$= 16,120$

Hope this helps.

6 0

3 years ago

Anyone who helps me answer I’ll brainliest them :D

Gala2k [10]

Answer:

5x -11

Step-by-step explanation:

it says it there

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3 years ago

Read 2 more answers

Can someone help me? Plz will mark brainliest!

tiny-mole [99]

1. - 3, +4 2. +2, - 5 3. +9, +1

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4 years ago

The harbormaster decides how many trips the ferry needs to make for 37 cars. The ferry can carry 8 cars at a time. What is the b

katen-ka-za [31]

Answer:

By adding 1 to the quotient.

Step-by-step explanation:

The Harbormaster needs to arrange tripe for 37 cars

in one go ferry can carry a maximum of 8 cars

trips = 37/8 = 4.625 trips

Since the ferry can't go in decimals so we need one more trip to carry all the cars. Hence the best way to interpret the remainder of 37/8 is by adding 1 to the quotient which will make 5 trips.

5 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

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

pounds of salt.

7 0

2 years ago