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anyanavicka [17]

3 years ago

14

Factor 12xy - 36xz a. 12(xy - 3xz) b. 12x(y - 3z) c. -24xyz d. 12(y - 3z)

1 answer:

weqwewe [10]3 years ago

6 0

Answer:

B is the correct answer.

Step-by-step explanation:

12xy - 36xz

12 and x are common factors.

12x(y - 3z)

Hope it helps.

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Perrys Bikes rents bikes for $17 plus $6 per hour. john paid $59 to rent a bike. For how many hours did he rent the bike?

matrenka [14]

7 hours

Let x represent hours rented
59=17+6x
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andrezito [222]

Answer:

D.

Step-by-step explanation:

In order to eliminate, the numbers must eliminate. If the second equation is multiplied by 3, then the 3 and the -3 can be eliminated. Since it is multiple choice, we can plug in x = 12 and y = 10 to see if they are correct. 12 + 30 = 42 and 24 - 10 = 14, so D is correct.

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A new health drink has 110% of the recommended daily allowance (RDA) for a certain vitamin. The RDA for this vitamin is 30 mg. H

ICE Princess25 [194]

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4 0

3 years ago

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What 10+1/3y=1 equal to

stepladder [879]

Answer:

y=11/3

Step-by-step explanation:

just 10 plus 1 gives 11 and crisscrossing and will get 3y=11 and dividing both sides by 3 to get y free then will get y=11/3

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3 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