I have 55 candy canes how many houses do i have to visit until i get 62 candy canes

Angelina_Jolie [31]

2х3+10х+2у2-х-у=2х3+9х+2у2-у=2*3in3+9*3+2*5in2-5=2*27+27+2*25-5=54+27+50-5=126

5 0

3 years ago

What is the range of the function g(x) = 3x^2 - 6x + 3 when the domain is defined as the set of integers, x, such that 0<=x&l

kakasveta [241]

Answer:

Range is 3 <= g(x) <= 27.

Step-by-step explanation:

The range is the values of g(x) for the given domain.

When x = 0 g(x) = 3(0)^2 - 6(0) + 3 = 3.

When x = 4 g(x) = 3(4)^2 - 6(4) + 3 = 27.

6 0

3 years ago

I have a strange True or False question on my packet.. But here I go: "A trapezoid has two 90 degree angles." Is it true or fals

alex41 [277]

Answer:

False unless a right trapezoid.

Step-by-step explanation:

All trapezoids have 4 angles which add to 360 degrees. The angles can be any measure except in a right trapezoid which does have two 90 degree angles. It is not a requirement to be a trapezoid though so this is false.

8 0

3 years ago

Layla used 0.482 gram of salt in her experiment. Maurice used 0.51 gram of salt in his experiment. Which pf the following amount

Sedaia [141]

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3 0

4 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