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valentinak56 [21]

3 years ago

12

Avenue A is perpendicular to North Street. What is the relationship between Avenue A and South Street?

1 answer:

MAVERICK [17]3 years ago

6 0

I need more information

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Which of the following describe an angle with a vertex at E?

Marianna [84]

Answer

B., C.

Step-by-step explanation:

Since the 'E' in the middle

it represents that the angle forms at the vertex E

3 0

3 years ago

Read 2 more answers

How many minutes equal 4.8 days? Enter your answer in the box.

vitfil [10]

1hour=60min
24hour=?
24hour=1440 min

1d=1440 min
4.8d=?
4.8 days =6956 hope you like my explanation

6 0

3 years ago

What are the degree and leading coefficient of the polynomial of 12u3 -8 + 23u

irga5000 [103]

Answer:

Degree: 3

Leading Coefficient: 12

Step-by-step explanation:

Find the highest power to determine the degree, in this case it's 3. Then identify the term with the highest power (12u^3) to find the leading coefficient.

So, because 3 is the largest and only power in this equation, it is the degree. 12u^3 is the leading term (term containing the highest degree) and 12 is the coefficient attatched to the term. Hope that helped!

4 0

3 years ago

Problem 10: A tank initially contains a solution of 10 pounds of salt in 60 gallons of water. Water with 1/2 pound of salt per g

AysviL [449]

Answer:

The quantity of salt at time t is $m_{salt} = (60)\cdot (30 - 29.833\cdot e^{-\frac{t}{10} })$ , where t is measured in minutes.

Step-by-step explanation:

The law of mass conservation for control volume indicates that:

$\dot m_{in} - \dot m_{out} = \left(\frac{dm}{dt} \right)_{CV}$

Where mass flow is the product of salt concentration and water volume flow.

The model of the tank according to the statement is:

$(0.5\,\frac{pd}{gal} )\cdot \left(6\,\frac{gal}{min} \right) - c\cdot \left(6\,\frac{gal}{min} \right) = V\cdot \frac{dc}{dt}$

Where:

$c$ - The salt concentration in the tank, as well at the exit of the tank, measured in $\frac{pd}{gal}$ .

$\frac{dc}{dt}$ - Concentration rate of change in the tank, measured in $\frac{pd}{min}$ .

$V$ - Volume of the tank, measured in gallons.

The following first-order linear non-homogeneous differential equation is found:

$V \cdot \frac{dc}{dt} + 6\cdot c = 3$

$60\cdot \frac{dc}{dt} + 6\cdot c = 3$

$\frac{dc}{dt} + \frac{1}{10}\cdot c = 3$

This equation is solved as follows:

$e^{\frac{t}{10} }\cdot \left(\frac{dc}{dt} +\frac{1}{10} \cdot c \right) = 3 \cdot e^{\frac{t}{10} }$

$\frac{d}{dt}\left(e^{\frac{t}{10}}\cdot c\right) = 3\cdot e^{\frac{t}{10} }$

$e^{\frac{t}{10} }\cdot c = 3 \cdot \int {e^{\frac{t}{10} }} \, dt$

$e^{\frac{t}{10} }\cdot c = 30\cdot e^{\frac{t}{10} } + C$

$c = 30 + C\cdot e^{-\frac{t}{10} }$

The initial concentration in the tank is:

$c_{o} = \frac{10\,pd}{60\,gal}$

$c_{o} = 0.167\,\frac{pd}{gal}$

Now, the integration constant is:

$0.167 = 30 + C$

$C = -29.833$

The solution of the differential equation is:

$c(t) = 30 - 29.833\cdot e^{-\frac{t}{10} }$

Now, the quantity of salt at time t is:

$m_{salt} = V_{tank}\cdot c(t)$

$m_{salt} = (60)\cdot (30 - 29.833\cdot e^{-\frac{t}{10} })$

Where t is measured in minutes.

7 0

3 years ago

2|4x-1|=38 I need to solve this

umka21 [38]

First you have to have the absolute value isolated so the 2 has to be divided. Then set up the two cases and solve.

5 0

3 years ago