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snow_lady [41]
3 years ago
6

Can someone help please!!! Pictures below

Mathematics
1 answer:
harkovskaia [24]3 years ago
8 0

Answer:

1.  im not really sure abt this one

<em>2. 11x-16</em>

3. 35+1/x+1

Step-by-step explanation:

<em>2. x^2/x^2=x</em>

<em>3X--7X=10X</em>

<em>-28+12=-16</em>

<em>x+10x=11x</em>

3. 5/1*6/1=35

35+1/x+1

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A bag contained 5 red, 8 blue, and 7 white marbles. Jose drew two marbles from the bag one at a time, without replacement. What
dezoksy [38]

Answer:

the probability that jose chose two blue marbles would be 8/20

Step-by-step explanation:

6 0
2 years ago
Solve the inequality below: <br><br> -2(x-5)&lt;4
alexandr402 [8]

Answer:

x > 3

Step-by-step explanation:

-2(x-5) < 4

-2x+10 < 4

-2x < -6

x > 3

7 0
3 years ago
In your own words, define each of the following terms. 
Leya [2.2K]
A. Whole number- a number that is whole, with no decimals or extra signs or variables next to or part of it. 

b. Digit- a number from 0-9 is a digit. 

c. The position a number has in a larger number. ex. Hundreds, Tens, and One's place

d. A number rounded to a whole number. Either to a smaller or larger to make it whole and easier to understand. 

e. Equals: Numbers that equal each other. 

I hope this helps you.

Brainliest answer would be appreciated!

6 0
3 years ago
Read 2 more answers
Don’t solve just write the equation!!!
erastovalidia [21]

Answer:

2(x - 4) = 16

8 0
2 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
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