Ask question

Ask question

All categories

3 years ago

15

Is 9r+16=π/5 a literal equation?

2 answers:

erik [133]3 years ago

6 0

The given equation is

$\rightarrow 9r+16=\frac{\pi}{5}$

The value of π is 3.14 or $\frac{22}{7}$ that is a fixed Value.

$\rightarrow 9r +16=\frac{3.14}{5}\\\\9r+16=0.628\\\\9r=0.628-16\\\\9r=-15.372\\\\r=\frac{-15.372}{9}\\\\r=-1.708$

So, the given equation is Literal ,because variable value is Known.

adell [148]3 years ago

4 0

That is not a literal equation, because the pi symbol is not a variable, you can do 3.14 divided by 5.

You might be interested in

Last season, a volleyball team won 40\%40% of 8080 games played.

IRISSAK [1]

Answer:

Happy Chrismas friend!

8 0

3 years ago

Initially a tank contains 10 liters of pure water. Brine of unknown (but constant) concentration of salt is flowing in at 1 lite

zhenek [66]

Answer:

Therefore the concentration of salt in the incoming brine is 1.73 g/L.

Step-by-step explanation:

Here the amount of incoming and outgoing of water are equal. Then the amount of water in the tank remain same = 10 liters.

Let the concentration of salt be a gram/L

Let the amount salt in the tank at any time t be Q(t).

$\frac{dQ}{dt} =\textrm {incoming rate - outgoing rate}$

Incoming rate = (a g/L)×(1 L/min)

=a g/min

The concentration of salt in the tank at any time t is = $\frac{Q(t)}{10}$ g/L

Outgoing rate = $(\frac{Q(t)}{10} g/L)(1 L/ min)$ $\frac{Q(t)}{10} g/min$

$\frac{dQ}{dt} = a- \frac{Q(t)}{10}$

$\Rightarrow \frac{dQ}{10a-Q(t)} =\frac{1}{10} dt$

Integrating both sides

$\Rightarrow \int \frac{dQ}{10a-Q(t)} =\int\frac{1}{10} dt$

$\Rightarrow -log|10a-Q(t)|=\frac{1}{10} t +c$ [ where c arbitrary constant]

Initial condition when t= 20 , Q(t)= 15 gram

$\Rightarrow -log|10a-15|=\frac{1}{10}\times 20 +c$

$\Rightarrow -log|10a-15|-2=c$

Therefore ,

$-log|10a-Q(t)|=\frac{1}{10} t -log|10a-15|-2$ .......(1)

In the starting time t=0 and Q(t)=0

Putting t=0 and Q(t)=0 in equation (1) we get

$- log|10a|= -log|10a-15| -2$

$\Rightarrow- log|10a|+log|10a-15|= -2$

$\Rightarrow log|\frac{10a-15}{10a}|= -2$

$\Rightarrow |\frac{10a-15}{10a}|=e ^{-2}$

$\Rightarrow 1-\frac{15}{10a} =e^{-2}$

$\Rightarrow \frac{15}{10a} =1-e^{-2}$

$\Rightarrow \frac{3}{2a} =1-e^{-2}$

$\Rightarrow2a= \frac{3}{1-e^{-2}}$

$\Rightarrow a = 1.73$

Therefore the concentration of salt in the incoming brine is 1.73 g/L

8 0

3 years ago

a bag contains 8 red 4 blue and 4 yellow marbles what is the probrability of randomly choosing 2 blue marbles without replacemen

enyata [817]

Answer:

12.5%

Step-by-step explanation:

8+4+4=16

2/16*100=12.5

12.5%

8 0

4 years ago

Emma works in a department store selling

Monica [59]

Answer:253.33...

Step-by-step explanation:

4 0

4 years ago

Can someone show the answer? need this quick :(

Oksanka [162]

Answer:

f(1) is 1, just look at the x-value provided in the parenthesis. f(1) means function value at x = 1, and the graph shows that at x = 1, y = 1.

8 0

3 years ago