Ask question

Ask question

All categories

2 years ago

5

What is 92/93 in the lowest terms

1 answer:

topjm [15]2 years ago

4 0

since 92 is not divisible into 93 by a whole number ratio, 92/93 is already in lowest terms

You might be interested in

Find the slope of the line passing through the points (-3,7) AND (2,-6)

castortr0y [4]

-13/5. The equation to find the slope between 2 points is y sub 2-y sub 1/x sub 2-x sub 1. Or in this case, -6-7/2+3. -6-7=-13, 2+3=5. So the slope is -13/5.

3 0

3 years ago

Read 2 more answers

5x + 16 = 9x-4 plz solve

RSB [31]

*Answer:*
x = 5

*step-by-step solution*

5 0

3 years ago

Read 2 more answers

Write 192.07 in words

Pavel [41]

Answer:

one hundred and ninety two point zero seven

Step-by-step explanation:

idk guess

7 0

3 years ago

Read 2 more answers

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

The cash price of a bike is $3000. The bike can be bought on hire purchase by making a 25% down-payment and monthly payments of

horrorfan [7]

25% downpayment of $750 + $125 x 24 months

= $3750

8 0

3 years ago

Read 2 more answers