Will mark brainliness, im really behind and trynna catch up :))

Find the sum of two consecutive numbers if the smaller one is 43

Alexxx [7]

Answer:

87

Step-by-step explanation:

Smaller number = 43

Bigger number= 44

Sum = 43+44

8 0

3 years ago

Read 2 more answers

Esteban is reading a book with 16 pages. He has read 9 pages so far. How many pages does Esteban have left to read? Write two eq

Zielflug [23.3K]

Answer:

7 pages

Step-by-step explanation:

Given the total number of pages in the book are 16

The number of pages esteban already read are 9

Let the number of pages left out to read be x

Total number of pages = number of pages read + number of pages to be read

16=9+x

x=16-9

x=7 pages

Therefore There are 7 pages left out to be read.

8 0

3 years ago

What is?<br><img src="https://tex.z-dn.net/?f=%20%5Cfrac%7Bx%7D%7By%7D%20%20%2B%20x" id="TexFormula1" title=" \frac{x}{y} + x"

Fiesta28 [93]

Answer:

$\frac{x+xy}{y}$

Step-by-step explanation:

Given

$\frac{x}{y}$ + x

Multiply x by $\frac{y}{y}$ to create a common denominator

= $\frac{x}{y}$ + x × $\frac{y}{y}$

= $\frac{x}{y}$ + $\frac{xy}{y}$

= $\frac{x+xy}{y}$

7 0

3 years ago

Help me plz Will be marked brainliest

Dmitriy789 [7]

Answer:

The answer is obviously letter A!

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