Ask question

Ask question

All categories

3 years ago

15

Please help! Kylie is 231 miles away from Anna. They are traveling towards each other. If Anna travels 9 mph faster than Kylie a

nd they meet after 7 hours, how fast was each traveling?

1 answer:

Andrej [43]3 years ago

7 0

A will represent Anna. k will represent Kylie.

7a+7k=231
a=9+k

Substitute for a
7(9+k)+7k=231
63+7k+7k=231
14k=168
k=12

Then plug in the k value
a=9+12
a=21

Final answer: Anna=21 mph, Kylie= 12 mph

You might be interested in

Two is subtracted from a number, and then the difference is multiplied by 5. The result is 30

Vanyuwa [196]

Answer:

8

Step-by-step explanation:

8-2=6

6x5=30

6 0

2 years ago

Read 2 more answers

Mrs. Haley baked apple pies for her family. The boys ate 9/10 of a pie and the girls ate 2/5 of a pie. How much more pie did the

garik1379 [7]

Answer: The boys ate half a pie more than the girls did (5/10 more).

Step-by-step explanation:

2/5 = 4/10

The difference between 9/10 and 4/10 is 5/10, or one half, so that's your answer.

4 0

3 years ago

What jacksepticeye's real name?

Alex Ar [27]

Answer:

Here is his real name Seán William McLoughlin

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Roger has a 0.250 batting average. If he went up to bat 240 times,how many times did he fail to get any hits?( A 0.250 batting a

gulaghasi [49]

Answer:

lol i need help with the same question

Step-by-step explanation:

7 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

2 years ago