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3 years ago

I need another opinion ΔΔΔΔ

Mathematics

2 answers:

Cloud [144]3 years ago

5 0

The answer should be c hope this helps

Svetllana [295]3 years ago

3 0

Yeah its c
cause they arent the same

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Find an equation of the line containing the given pair of points (5,2) and (-3,5)

anzhelika [568]

(5,2)(-3,5)
slope = (5 - 2) / (-3 - 5) = -3/8

y = mx + b
slope(m) = -3/8
use either of ur points...(5,2)...x = 5 and y = 2
now we sub and find b, the y int
2 = -3/8(5) + b
2 = - 15/8 + b
2 + 15/8 = b
16/8 + 15/8 = b
31/8 = b

so ur equation is : y = -3/8x + 31/8....or 3x + 8y = 31

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3 years ago

Determining slope and y-intercept

maw [93]

Answer:

The slope is -1

Step-by-step explanation:

Anytime the x-intercept values have 0 in it, the y-intercept value which is below/above the x-values (Which in this case is -1) is the y-intercept.

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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 can you explain the answer 4 divided by 5

Sergio [31]

Answer:

4 divided by 5 is equal to 0.8. This decimal can also be written as a fraction. 0.8 = eight tenths or 8/10 (4/5 in its reduced form).

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Given: BC bisects DBE. Prove: ABD is congruent to ABE

Arada [10]

Answer:

See below

Step-by-step explanation:

BC bisects <DBE and if AC is a straight line then you have that:

<ABD + <DBC = 180 (straight angle because of line AC)

<ABE + <CBE = 180 ( straight angle because of line AC)

Because BC bisects <DBE => < DBC = <CBE

So <ABD and <ABE must be the same to both sum 180 when added < DBC

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3 years ago