All categories

3 years ago

What is the algebraic expression for the following word phrase: the quotient of j and 8

Mathematics

2 answers:

DanielleElmas [232]3 years ago

8 0

Answer:

$\frac{j}{8}$

Step-by-step explanation:

the quotient of j and 8

We need to write the given word phrase in algebraic expression

Algebraic expression contains terms with numbers and variables with operators in between.

In the given word phrase , 'j' is the variable

quotient occurs only on division . so we divide 'j' and 8

The expression for word phrase the quotient of j and 8 is

$\frac{j}{8}$

kirill [66]3 years ago

5 0

$\frac{j}{8}$

This is the correct algebraic expression for the quotient of j and 8. Quotient is just a fancy term for the answer to a division problem.

You might be interested in

Estimate. Then find the difference 300,980 -159,000

Softa [21]

Answer:

142,000

Step-by-step explanation:

I would round 300,980 to the nearest thousands place = 301,000

Subtract 301,000 and 159,000 = 142,000

4 0

3 years ago

Fill in the missing term and select the missing description. Simplify any fractions.

mina [271]

Answer:

t = 18/2

= 9

Sorry only thing I know is this!!

6 0

3 years ago

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