Ask question

Ask question

All categories

4 years ago

9

Larissa has 4 ? cups of flour. She is making cookies using a recipe that calls for 2 ? cups of flour. After baking the cookies,

how much flour will be left?

1 answer:

Scorpion4ik [409]4 years ago

4 0

Answer:

2 cups of flour

Step-by-step explanation:

4-2 = 2

You might be interested in

If y = 5x − 4, which of the following sets represents possible inputs and outputs of the function represented as ordered pairs?

zzz [600]

The answer is not in the choices so ... but the (0,4), (1,1), (2,6)

Hope This Help.

Xd <3

6 0

3 years ago

Which is the best estimate 9.6-6.2

velikii [3]

The best estimate is probably 4

6 0

3 years ago

Read 2 more answers

the volume of a rectangular solid, in cubic inches, is given by the expression (x^3-7x+6). if the length and width in inches, ar

nekit [7.7K]

V=x³-7x+6, V=(x-2)(x+3)*h

(x-2)(x+3)*h=x³-7x+6

h{x²+x-6}=x³-7x+6

h=(x³-7x+6)/(x²+x-6)

Therefore h=x-1

-----------------------------

Check answer:

V=(x-2)(x+3)(x-1)

=(x-1)(x²+x-6)

=x(x²+x-6)-1(x²+x-6)

=x³+x²-6x-x²-x+6

=x³-6x-x+6

=x³-7x+6

5 0

3 years ago

Round your answers to one decimal place, if necessary.

Marina CMI [18]

Answer: 11

Step-by-step explanation:

Add the numbers together to get 68, then divide it by 6 to get your answer! Sorry if I'm wrong!

5 0

3 years ago

Read 2 more answers

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