HELP YA GIRL PLZZZZ I WILL REWARD

I WILL AWARD BRAINLIEST!! PLEASE HELP!!!<br> Find the areas of the trapezoids.

Galina-37 [17]

Answer:

A = 12 units ^2

Step-by-step explanation:

The area of the trapezoid is found by

A = 1/2 (b1+b2)h

b1 = 2

b2 = 4

h = 4

I found these by looking at the graph

A = 1/2(2+4) 4

A = 1/2(6*4)

A = 12 units ^2

8 0

2 years ago

Hello i hope you have a great day (:

Vlad [161]

Answer:

Ty, you too!

Step-by-step explanation:

Hope you have a good day.

7 0

2 years ago

Read 2 more answers

How do u rewrite 0.72 repeating as a simplified fraction?

vampirchik [111]

8/11 is the answer

I use this trick: When tryig to find repeating numbers divide by 11 or 9

When dividing by 9 the numerator will be the first number in the decimal

When dividing by 11 the numerator will be one more than the first number in the decimal

8 0

2 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

How much money must you pay into an account at the beginning of each of 20 years in order to have $10,000 at the end of the 20th

Blababa [14]

Answer:

$191

Step-by-step explanation:

6 0

2 years ago