All categories

3 years ago

Find the quotient if the dividend is 2.26 · 108 and the divisor is 0.013.

Mathematics

1 answer:

Sliva [168]3 years ago

5 0

Answer:

17.38*10^8

Step-by-step explanation:

Step one:

given data

The dividend is 2.26*10^8 (i rewrote the given data, i hope it was mearnt to be so)

divisor is 0.013.

Step two:

Hence the quotient is the whole number gotten when the dividend is divided by the divisor

quotient= 2.26*10^8/0.013.

quotient= 17.38*10^8

You might be interested in

Statement Reason

andrew11 [14]

Answer:vhv

Step-by-step explanation:hhgjhghjgjg

7 0

3 years ago

Brad wants to buy flowers for his friend with $15. The daisies are $1 each and the roses are $3 each. He buys 3 more daisies tha

melisa1 [442]

D + 3r = 15
d = r + 3

r + 3 + 3r = 15
4r + 3 = 15
4r = 15 - 3
4r = 12
r = 12/4
r = 3 ....he bought 3 roses, at $ 3 per rose = $ 9 <==

d = r + 3
d = 3 + 3
d = 6....he bought 6 daisies, at $ 1 per daisy = $ 6

6 0

3 years ago

Read 2 more answers

Hello guys i m new here vkq-jnfi-uif meet

sergeinik [125]

Answer:

welcome

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Please help on the production one.

kirza4 [7]

5x + 8000 = 3x + 10000
x = 1000

5 0

3 years ago

Suppose water is added to a tank at 10 gal / min , but leakes out at the rate of 0.2 gal / min for each gallon in the tank . Wha

Nana76 [90]

Answer:

50 gallon

Step-by-step explanation:

We are given that

$\frac{dV_{in}}{dt}=10 gal/min$

Let V(in gallon) be the present volume of tank.

$\frac{dV_{out}}{dt}=0.2 V gal/min$

We have to find the smallest capacity the tank can have if the process is to continue indefinitely.

According to question

$\frac{dV}{dt}$ =Incoming volume-outgoing volume

$\frac{dV}{dt}=10-0.2 V=-(\frac{1}{5}V-10)=-(\frac{V-50}{5})$

$\frac{5dV}{V-50}=-dt$

Integrating on both sides

$\int \frac{5dV}{V-50}=-\int dt$

$5ln(V-50)=-t+ln C$

By using the formula

$\int \frac{dx}{x}=ln x+C$

$ln(V-50)=\frac{1}{5}(-t+ln C)$

$ln(V-50)=-\frac{1}{5}t+\frac{1}{5} ln C$

$ln(V-50)=-\frac{1}{5} t+lnC^{\frac{1}{5}}$

$ln(V-50)=-\frac{1}{5}t+ln C'$

Where $ln C'=ln C^{\frac{1}{5}}$

$ln(V-50)-ln C'=-\frac{1}{5} t$

$ln\frac{V-50}{C'}=-\frac{1}{5}$

Using the formula

$ln m-ln n=ln\frac{m}{n}$

$\frac{V-50}{C'}=e^{-\frac{1}{5}t}$

$V-50=C'e^{-\frac{1}{5}t}$

Substitute $t=\infty$

$V-50=0$

$V=50$

Hence, the smallest capacity the tank can have if the process is to continue indefinitely=50 gallon

7 0

3 years ago