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3 years ago

8

I don’t understand can someone please help me

1 answer:

alexandr1967 [171]3 years ago

4 0

Answer:

Answer:

Step-by-step explanation:

Answer as Photo

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What is Evaluate<br>120 + 5

NeX [460]

Answer:

24

Step-by-step explanation:

1: Divide 120 by 5 = 24

Answer: 24

<em><u>Hope this helps.</u></em>

7 0

3 years ago

5. Here is a rectangular prism.

vovangra [49]

Answer:

28.8sq area is the area of the prism

Step-by-step explanation:

2.4x1.5x8

7 0

2 years ago

Helpppppppppppppppp ill mark you brainlist write in y=mx+b form

Alex Ar [27]

Answer:

I think: 0 = 3x + b or b = -3x + 0 or x = -b + 0/3 or 3 = -b + 0/x

Sorry if you get this wrong..

4 0

3 years ago

Following the lead of the National Wildlife Federation, the Department of the Interior of a South American country began to reco

chubhunter [2.5K]

Answer:

After 2009 the quality of the forests in this country improving.

Step-by-step explanation:

Consider the provided function.

$f(t) = \frac{1}{3}t^3-\frac{5}{2}t^2 + 80$

$f'(t) = t^2-5t$

Substitute f'(t)=0.

$t^2-5t=0$

$t(t-5)=0$

$t=0\ or\ t=5$

It is given that (0 ≤ t ≤ 10)

The test interval are (0,5) and (5,10)

For (0 < t < 5) the value of $f'(t)$

For (5 < t < 10) the value of $f'(t) >0$

Hence, the function is increasing from (5,10).

t = 0 corresponds to 2004.

Therefore, after 2009 the quality of the forests in this country improving.

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