Given: 3x + y = 1 Solve for y. A. y=-3x + 1 B. y=-3x - 1 C. y= 3x

How to graph the line y=4/3x

ankoles [38]

Answer:

make a table of values

Step-by-step explanation:

then plot using those values

3 0

3 years ago

What is the exact circumference of the circle?

Arlecino [84]

I am pretty sure it is 180 degrees

5 0

3 years ago

Read 2 more answers

As – 16a, as – 4a3, 2a3 + 8a - 4a2 - 16

vfiekz [6]

Answer:

1) as-16a

taknig a as common

=a(s-16)

2)as-4a³

taking a as common

=a(s-4a²)

3)2a³+8a-4a²-16

taking common terms

=2a(a²+4)-4(a²+4)

=(2a-4)(a²+4)

again taking common

=2(a-2)(a²+4)

i hope this will help you :)

7 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

how many square feet will cover a rectangle 60 feet by 20 feet? Explain how you determined your answer

nika2105 [10]

Answer:

1200

Step-by-step explanation:

2 times 6 = 12

12 times 100 = 1200

Hope this helps if does not please let me know thanks!

7 0

2 years ago

Given: 3x + y = 1 Solve for y. A. y=-3x + 1 B. y=-3x - 1 C. y= 3x - 1