1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Tpy6a [65]
3 years ago
5

The function h(t) = -4.9t² + 19.6t is used to model the height of an object projected in the air where h(t) is the height (in me

ters) and t is the time (in seconds). What is the domain and range? Domain:
Mathematics
2 answers:
goblinko [34]3 years ago
5 0

Answer:

Step-by-step explanation:

when h(t)=0

-4.9 t²+19.6t=0

4.9t(-t+4)=0

either t=0 or t=4

so domain is 0≤t≤4

for range

h(t)=-4.9t²+19.6t

=-4.9(t²-4t+4-4)

=-4.9(t-2)²+19.6

so range is 0≤h≤19.6

snow_tiger [21]3 years ago
3 0

Domain = 0<t<4, make sure to use less than or equal to signs not just less than signs.

Range = 0<h<19.6, again, use less than or equal to signs.

You might be interested in
What is the percent of games won if a team won 92 games and lost 58 games?
Lera25 [3.4K]

Answer: 59.38%

there is the answer

5 0
3 years ago
Read 2 more answers
WHAT IS 9 x 99 x 999 x 9999 x 99999 x 999999 x 9999999 x 99999999 x 999999999 x 9999999999 &gt;:)
Schach [20]

Answer:

8.900101E54

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
May someone do this for me please?
Nesterboy [21]
58 should be the answer for the perimeter
8 0
3 years ago
Read 2 more answers
Two cars leave the same location traveling in opposite directions. One car leaves at 3:00 p.m. traveling at an average rate of 5
sergey [27]
Recall your d = rt,   distance = rate * time

thus \bf \begin{array}{lccclll}&#10;&distance&rate&time(hrs)\\&#10;&-----&-----&-----\\&#10;\textit{first car}&d&55&x\\&#10;\textit{second car}&380-d&75&x+1&#10;\end{array}\\\\&#10;-----------------------------\\\\&#10;&#10;\begin{cases}&#10;\boxed{d}=(55)(x)\\\\&#10;380-d=(75)(x+1)\\&#10;----------\\&#10;380-\left( \boxed{(55)(x)} \right)=(75)(x+1)&#10;\end{cases}

notice, the first car leaves at "x" time, the other leaves on hour later, or x + 1

the first car travels some distance "d", whatever that is, thus
the second car, picks up the slack, or the difference, they're 380 miles
apart, thus the difference is 380-d
8 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
Other questions:
  • Wilma and Betty are playing a number game. Wilma tells Betty, "I'm thinking of a number between 1 and 10. If I take the number a
    9·1 answer
  • Olivia walks 3 1/2 miles per hour. how far can she walk in 2 1/4 hours?
    10·1 answer
  • An equation parallel and perpendicular to 4x+5y=19
    13·1 answer
  • Anyone care to help? :)
    11·1 answer
  • According to 2015 census data, 42.7 percent of Colorado residents were born in Colorado. If a sample of 250 Colorado residents i
    15·1 answer
  • SOS HURRY
    9·1 answer
  • Solve the system of equations please<br> 1/4x+y=1<br> 3/2x-y=4/3
    15·1 answer
  • Simplify the expiration 2h+6h+4h
    6·1 answer
  • Stuck On this one too. Think I’m missing a step
    5·1 answer
  • Will give <br> BRAINLY AND FIVE STARS and A THANKS
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!