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3 years ago

The highest point over the entire domain of a function or relation

Mathematics

1 answer:

mariarad [96]3 years ago

3 0

Answer:

The highest point over the entire domain of a function or relation is called an <u>absolute maximum.</u>

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Step-by-step explanation:

For a continuous function f(x) in a closed interval, there exists a value of f(x) that is the highest and that point is called the absolute maximum. The maximum value can be found a number of ways including a graphical method and using the first and second derivative.

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Find the measures of the angles given

dexar [7]

Answer:

BCD = 60

ABD - 90

Step-by-step explanation:

BCD)

BC and CD are the same length on in the circle, they are the radius. Therefor, all angles will be the same at 60

ABD)

All angles together are 180

You know 2 Angles, A at 30 and D at 60

180 - 30 - 60 = Angle B

90 = Angle B

6 0

3 years ago

I WILL GIVE 20 POINTS TO THOSE WHO ANSWER THIS QUESTION RIGHT NOOOO SCAMS AND EXPLAIN WHY THAT IS THE ANSWER

luda_lava [24]

Answer:

Step-by-step explanation:

7 0

2 years ago

8)Reduce a las unidades indicadas: a)18 cm 3 a dm 3

lys-0071 [83]

Answer:

(a) 0.018 dm^3

(b) 20 x 10^15 ml^3

Step-by-step explanation:

a) 18 cm^3 to dm^3

1 cm = 0.1 dm

So,

1 cm^3 = 0.001 dm^3

So, 18 cm^3 = 18 x 0.001 dm^3 = 0.018 dm^3

b) 20 hl^3 to ml^3

1 hl = 100000 ml

So, 1 hl^3 = 10^15 ml^3

So,

20 hl^3 = 20 x 10^15 ml^3.

7 0

3 years ago

Suppose that a large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved. Another br

aliya0001 [1]

Answer:

$A=1500-1450e^{-\dfrac{t}{250}}$

Step-by-step explanation:

The large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved.

Volume = 500 gallons

Initial Amount of Salt, A(0)=50 pounds

Brine solution with concentration of 2 lb/gal is pumped into the tank at a rate of 3 gal/min

$R_{in}$ =(concentration of salt in inflow)(input rate of brine)

$=(2\frac{lbs}{gal})( 3\frac{gal}{min})\\R_{in}=6\frac{lbs}{min}$

When the solution is well stirred, it is then pumped out at a slower rate of 2 gal/min.

Concentration c(t) of the salt in the tank at time t

Concentration, $C(t)=\dfrac{Amount}{Volume}=\dfrac{A(t)}{500}$

$R_{out}$ =(concentration of salt in outflow)(output rate of brine)

$=(\frac{A(t)}{500})( 2\frac{gal}{min})\\R_{out}=\dfrac{A}{250}$

Now, the rate of change of the amount of salt in the tank

$\dfrac{dA}{dt}=R_{in}-R_{out}$

$\dfrac{dA}{dt}=6-\dfrac{A}{250}$

We solve the resulting differential equation by separation of variables.

$\dfrac{dA}{dt}+\dfrac{A}{250}=6\\$The integrating factor: e^{\int \frac{1}{250}dt} =e^{\frac{t}{250}}\\$Multiplying by the integrating factor all through\\\dfrac{dA}{dt}e^{\frac{t}{250}}+\dfrac{A}{250}e^{\frac{t}{250}}=6e^{\frac{t}{250}}\\(Ae^{\frac{t}{250}})'=6e^{\frac{t}{250}}$

Taking the integral of both sides

$\int(Ae^{\frac{t}{250}})'=\int 6e^{\frac{t}{250}} dt\\Ae^{\frac{t}{250}}=6*250e^{\frac{t}{250}}+C, $(C a constant of integration)\\Ae^{\frac{t}{250}}=1500e^{\frac{t}{250}}+C\\$Divide all through by e^{\frac{t}{250}}\\A(t)=1500+Ce^{-\frac{t}{250}}$

Recall that when t=0, A(t)=50 (our initial condition)

$50=1500+Ce^{-\frac{0}{250}}50=1500+Ce^{0}\\C=-1450\\$Therefore the amount of salt in the tank at any time t is:\\A=1500-1450e^{-\dfrac{t}{250}}$

4 0

3 years ago

A hiker walks at a speed of 3 2 5 kilometers per hour. How many kilometers will the hiker walk in

Vika [28.1K]

Answer:

= 2 4/15 km

Step-by-step explanation:

Speed = 3 2/5 km/hour

The units are not the same. The speeds is in hours and the time is in minutes

Lets convert minutes to hours. 60 minutes = 1 hour

40 minutes * 1 hour/ 60 minutes = 40/60 hours = 2/3 hour

Distance = speed * time

= 3 2/5 * 2/3

Change the mixed number to an improper fraction

3 2/5 = (5*3+2)/5 = 17/5

= 17/5 * 2/3

= 34/ 15 km

Change this back to a mixed number

15 goes into 34 2 times (15*2 = 30) with 4 left over

= 2 4/15 km

3 0

4 years ago