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vladimir1956 [14]

3 years ago

14

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1 answer:

elena55 [62]3 years ago

7 0

?!?! whaaatt confused lol

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Four whole numbers are each rounded to the nearest 10. The sum of the four rounded numbers is 90. What is the maximum possible s

qwelly [4]

Answer:Whats the highest number each number can go up to before it rounds up? Its +4 as each number will round up otherwise. So if the rounded up numbers were 10,20,30,30. They can go up to 14,24,34,34. So its 106.please mark as brainliest

Step-by-step explanation:

5 0

3 years ago

how would you arrange four 3's so they equal the number 0 through 10? I did 0 and 1 need help with the other numbers PLEASES!!!

Harlamova29_29 [7]

3+3+3-3=6

(3+3+3-3)÷2=3

(3+3+3+3)÷3=4

(3+3+3+3)÷6=2

Is this how it works?

5 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

The period of a pendulum is given by the equation

sattari [20]

Answer: g > 0

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

At the basketball game, Leo scored points in 30% of his attempts. Which of the following statements is true?

Nuetrik [128]

Answer:

C. Leo scored points in 3/10 of his attempts

Step-by-step explanation:

30% is 30/100

Which can simplify to

3/10

5 0

3 years ago