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3 years ago

13

Find the range of the relation.

1 answer:

svlad2 [7]3 years ago

5 0

D the last option ......

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On day 4 the 1rst 2 <br> thx

ladessa [460]

1. C
2. 3, because I used a proportion table

8 0

3 years ago

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What is a solution for 4a = 36?

ycow [4]

4a = 36

to solve for a, divide both sides by 4 to get a by itself.

<u>4a</u> = <u>36
</u> 4 4
<u>
</u>a = 9

8 0

2 years ago

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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

which of the following transformations does not preserve distance and angle measures? A .translation B.reflection C.rotation D.h

wel

<span>D.horizontal stretch
</span>

5 0

3 years ago

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2/3 + 1/2 in simple form fractions

nata0808 [166]

Try and find the common denominator which is 6 since 2×3 is 6 then once u get ur denominator u have to find the numerator which would be 4/6+3/6 And that would equal 7/6 and if you would u would keep it like that but if u would have to change it into a mixed fraction it would be 1 and 1/6

Hoped this helped

3 0

3 years ago