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2 years ago

10

What's the answer ? To this problem

2 answers:

Nataliya [291]2 years ago

3 0

The answer is A. You can solve this by plugging the X values from the table into the equation and seeing if you end up with the right Y value.

Liula [17]2 years ago

3 0

The answer is A do you need the steps ?

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Which expression is equivalent to 1/3(4x+9)+5x?

puteri [66]

I think it d. I’m not sure I could be wrong

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2 years ago

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3 lb 9oz + 5lb 10 oz.

Eduardwww [97]

Answer:

9 lb 3 oz

Step-by-step explanation:

(3 + 5) lb + (9 + 10) oz

8 lb + 19 oz

16 oz = 1 lb

(8 + 1) lb + (19 - 16) ox

9 lb 3 oz

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2 years ago

Describe a time when someone around you made a decision based on authority of someone else?

dalvyx [7]

There was a time that a friend of mine got to a decision because the leader of the group said so (it was a group project). Although it was clear that he did not want to comply to the decision, he made it because it was what the leader said so.

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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

2 years ago

In the calm waters of an inlet, Manon's boat moves with a velocity (speed and direction) vector \vec{b_1} = (3,4) b 1 =(3,4)

andrew-mc [135]

Answer with Step-by-step explanation:

We are given that

$b_1=(3,4)$

$b_2=(1,1)$

We have to find the speed of river's current and the direction of current flowing.

$b=b_2-b_1=(1,1)-(3,4)=(-2,-3)$

Speed of river's current= $\mid b\mid$

Speed of river's current= $\sqrt{(-2)^2+(-3)^2}$

Speed of river's current= $\sqrt{13} m/s$

By using the formula

$\mid r\mid =\sqrt{x^2+y^2}$

Direction of current flowing, $\theta=tan^{-1}(\frac{y}{x})$

$\theta=tan^{-1}(\frac{-3}{-2})=56.3^{\circ}$

The angle lies in third quadrant therefore

$\theta=180+56.3=236.3^{\circ}$

3 0

2 years ago