Find the distance BF for the points B(1, –1) and F(

Calculate the gradient of the straight line which joins A(2,-3) and B(6,9)<br>

Pie

Answer:

Gradient: m = 3

Step-by-step explanation:

m = change in y/change in x

m = $\frac{Y2-Y1}{X2-X1}$

m = $\frac{9-(-3)}{6-(2)}$

m = 12/4

m = 3

5 0

3 years ago

What is the volume, in cubic centimeters, of a right rectangular prism that has a lenght of 4 centimeters, a width of 9 centimet

Temka [501]

Answer:

The volume of the prism is 360 cm^3.

Step-by-step explanation:

The rectangular prism is a three dimensional figure formed by six rectangular faces. It's volume is given by the product of it's three dimensions. The calculation for the volume of this prism is shown bellow:

volume = width*height*length

volume = 9*10*4

volume = 360 cm^3

The volume of the prism is 360 cm^3.

7 0

3 years ago

Read 2 more answers

Which equation applies the associative property of multiplication?

ohaa [14]

Answer:

5 0

2 years ago

In a flower garden, there are 3 tulips for every 8 daisies. If there are 18 tulips, how many daisies are there?

LenaWriter [7]

18/3 = 6
So 8*6 = 48
There will be 48 daisies

6 0

3 years ago

Read 2 more answers

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

Find the distance BF for the points B(1, –1) and F(–3, 5).