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3 years ago

Assuming x=0 and y=0, what is the quotient of 24^6y^2 + 16x^4y^3 + 8x^2y^4

2 answers:

6 0

There is no quotient: if x is equal to zero, then your problem would simplify to a number divided by zero. As it is impossible to divide by zero, there is no answer.

dsp733 years ago

6 0

There is no answer because anything to the power of zero is 1 and 0 to the power of anything is zero. So if the end result is 1/0 and nothing can divide into zero then there is no answer.<span />

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If AP 10 - 12. what is the perimeter of quadrilateral TAEO (a) 56 (b)64 (c)72 (d)76

BlackZzzverrR [31]

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4 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

There are 16 teams in a tournament. if during the first round, each team plays every other team exactly once, how many games wil

jasenka [17]

16x17=272/2=136. That is answer.

5 0

3 years ago

bag contains 6 red marbles and 12 yellow marbles. If a representative sample contains 3 red marbles, then how many marbles woul

Dafna11 [192]

I think 6 because it’s a ratio

Because the yellow would always be twice as much as the red so 3x2 is 6 just like 6x2=12

7 0

3 years ago

The revenue from selling x shirts is r(x) = 11x. The cost of buying x shirts is c(x) = 6x + 20. The profit from selling x shirts

oee [108]

Answer: option A is the correct answer.

Step-by-step explanation:

The expression that relates revenue, profit and cost is

Profit = Revenue - Cost

The revenue from selling x shirts is r(x) = 11x. The cost of buying x shirts is c(x) = 6x + 20. Therefore,

p(x) = r(x) - c(x)

p(x) = 11x - (6x + 20)

By expanding the bracket, the minus sign will alter each term in the bracket. Therefore,

p(x) = 11x - 6x - 20

p(x) = 5x - 20

8 0

2 years ago