Ray’s weight increased by 11% in the last two years. If he gained 16.5 pounds, what was his weight two years ago?

The distance of one lap around a school track is 1/4 miles. Karen walks 6 laps around the track and realizes it is in

Gemiola [76]

Answer:

16

Step-by-step explanation:

because 5.5 is 22 laps and you do 22-6=16.

7 0

3 years ago

Find the missing angle

Zanzabum

58 degrees……….:..:………..

3 0

3 years ago

Read 2 more answers

HELPPP MEEEE PLSSSS I WILL GIVE CROWN

emmasim [6.3K]

Answer:

6

Step-by-step explanation:

if you notice here Josiah starts with 250 and Katie starts with 215, and in Josiah's equation he saves 25$ each month while Katie save 45$ each month, in order to find our answer we must multiply there starting amount by how many months.

A: 4 months is incorrect because Katie has 45x4+215 is 335, and Josiah has 25x4+250 is 350, so Katie is still behind

B: 5 months is incorrect  because Katie has 45x5+215 is 375, and Josiah has, 25x5+250 which is 375, so they have the same amount of savings

C: 6 months is correct because Katie has 45x6+215 is 420, and Josiah has, 25x5+250 which is 400 , so Katie has more savings, Thus C is correct

Sorry for the Wait, Your Welcome, And Have An Amazing Day

3 0

3 years ago

14.

FromTheMoon [43]

Answer:

a) 200 yards

Step-by-step explanation:

Difference in the distance = 50 yards

Difference in speed = 8 - 6 = 2 y/s

Time to cover the gap is:

50/2 = 25 seconds

Sandra will run the distance:

25*8 = 200 yards

8 0

3 years ago

Read 2 more answers

A 1000-liter (L) tank contains 500 L of water with a salt concentration of 10 g/L. Water with a salt concentration of 50 g/L flo

djverab [1.8K]

Answer:

a) $y(t)=50000-49990e^{\frac{-2t}{25}}$

b) $31690.7 g/L$

Step-by-step explanation:

By definition, we have that the change rate of salt in the tank is $\frac{dy}{dt}=R_{i}-R_{o}$ , where $R_{i}$ is the rate of salt entering and $R_{o}$ is the rate of salt going outside.

Then we have, $R_{i}=80\frac{L}{min}*50\frac{g}{L}=4000\frac{g}{min}$ , and

$R_{o}=40\frac{L}{min}*\frac{y}{500} \frac{g}{L}=\frac{2y}{25}\frac{g}{min}$

So we obtain. $\frac{dy}{dt}=4000-\frac{2y}{25}$ , then

$\frac{dy}{dt}+\frac{2y}{25}=4000$ , and using the integrating factor $e^{\int {\frac{2}{25}} \, dt=e^{\frac{2t}{25}$ , therefore $(\frac{dy }{dt}+\frac{2y}{25}}=4000)e^{\frac{2t}{25}$ , we get $\frac{d}{dt}(y*e^{\frac{2t}{25}})= 4000 e^{\frac{2t}{25}$ , after integrating both sides $y*e^{\frac{2t}{25}}= 50000 e^{\frac{2t}{25}}+C$ , therefore $y(t)= 50000 +Ce^{\frac{-2t}{25}}$ , to find $C$ we know that the tank initially contains a salt concentration of 10 g/L, that means the initial conditions $y(0)=10$ , so $10= 50000+Ce^{\frac{-0*2}{25}}$

$10=50000+C\\C=10-50000=-49990$

Finally we can write an expression for the amount of salt in the tank at any time t, it is $y(t)=50000-49990e^{\frac{-2t}{25}}$

b) The tank will overflow due Rin>Rout, at a rate of $80 L/min-40L/min=40L/min$ , due we have 500 L to overflow $\frac{500L}{40L/min} =\frac{25}{2} min=t$ , so we can evualuate the expression of a) $y(25/2)=50000-49990e^{\frac{-2}{25}\frac{25}{2}}=50000-49990e^{-1}=31690.7$ , is the salt concentration when the tank overflows

4 0

4 years ago