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Dmitrij [34]
3 years ago
13

Determine where V(z)=z4(2z−8)3 is increasing and decreasing.

Mathematics
1 answer:
Marat540 [252]3 years ago
8 0

Answer and Explanation:

A function is said to be increasing, if the derivative of function is f’(x) > 0 on each point. A function is said to be decreasing if f”(x) < 0.

Let y = v (z) be differentiable on the interval (a, b). If two points z1 and z2 belongs to the interval (a, b) such that z1 < z2, then v (z1) ≤ v (z2), the function is increasing in this interval.

Similarly, the function y = v(z) is said to be decreasing, when it is differentiable on the interval (a , b).

Two points z1 and z2 Є (a, b) such that z1 > z2, then v (z1) ≥ v(z2). The function is decreasing on this interval.

The function y = v (z)

The derivative of function Y’ = v’(z) is positive, then the function is increasing.

The function y = v (z)  

The derivative of function y’ is negative, then the function is decreasing.

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Which of the following shows a true statement? -12 + (-7) = 12 - 7
Marizza181 [45]

Answer:

ok so its the first one

Step-by-step explanation:

it shows th same kind of aleration

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3 years ago
PLEASE ANSWER ASAP Which function has a constant additive rate of change of –1/4?
Hatshy [7]

Answer:

the first graph

Step-by-step explanation:

The first one  - y = 1/4x + 3/2.

sorry im a bit late but hopefully this helps

6 0
3 years ago
For each of the following numbers, write 4 decimals that would round to this whole number when rounde to the nearest whole numbe
Natalka [10]
Round to the nearest whole number:

a) 14 ⇒ 13.5 ; 13.9 ; 14.1 ; 14.4
b) 28 ⇒ 27.5 ; 27.9 ; 28.1 ; 28.4
c) 32 ⇒ 31.5 ; 31.9 ; 32.1 ; 32.4
d) 50 ⇒ 49.5 ; 49.9 ; 50.1 ; 50.4
e) 67 ⇒ 66.5 ; 66.9 ; 67.1 ; 67.4
f) 71 ⇒ 70.5; 70.9 ; 71.1 ; 71.4
g) 88 ⇒ 87.5; 87.9 ; 88.1 ; 88.4
h) 95 ⇒ 94.5; 94.9 ; 95.1 ; 95.4

If the following number is within the range of 1 to 4 then round down.
<span>If the following number is within the range of 5 to 9 then round up.</span>
8 0
3 years ago
Come hither big brains
katrin2010 [14]

Answer:

The answer is c

Step-by-step explanation:

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
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