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yarga [219]
3 years ago
15

How many 4 digit numbers can be made from 4760

Mathematics
2 answers:
Arturiano [62]3 years ago
4 0
You can do 0764 0476. 4670. 7640
irinina [24]3 years ago
3 0
7604 6047 7064 4670 are some options
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Can someone please help me, this is the last question on my test and I dont understand. please and thanks.
Elina [12.6K]

Step-by-step explanation:

A: ←⭕

B: •→

C: ←•

D: ⭕→

.........

4 0
2 years ago
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Find the value of the greater root of x^2 + 10x + 24x = 0 a) -6 b) -4 c) 4 d) 6
Lubov Fominskaja [6]

what formula is applied, I mean what method

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2 years ago
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Given m||n, find the value of x.<br><br> (7%-10)<br> (6X-5)
tamaranim1 [39]

15 degrees

7x-10+6x-5=180

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6 0
3 years ago
What is the average of -30 -15 -20 55
defon

Answer:

-2.5

Step-by-step explanation:

-30 + (-15) + (-20) + 55 = -10. Then you divide it by how many numbers there are (4). Which equals - 2.5. If this is wrong, then round up to -3.

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
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