Hello everyone (this is 6th grade math)

3.3.1. Calculate the new cost per night of the standard and family standard chalets. Round of (4) your answer to the nearest hun

Taya2010 [7]

The information to complete the table is given below:

Equation that shows the relationship between the cost per night C and the number of people staying in the chalet n: C= 4000-100(n-1)

<h3>What is Equation?</h3>

An equation is a mathematical expression that contains an equals symbol. Equations often contain algebra. Algebra is used in math when you do not know the exact number in a calculation.

Given that:

Pules Rs 4000 per night but will decrease the amount by R100 per person occupying the chalet.

Number of People Cost per night

1 4000

2 3900

3 3800

4 3700

5 3600

6 3500

7 3400

8 3300

9 3200

10 3100

let C be cost per night and n be the number of people of staying in chalet.

equation: C= 4000-100(n-1)

for n=1: C= 4000- 0=4000

n=2, C= 4000-100(1)= 3900

The graph is attached below:

Learn more about equation here:

brainly.com/question/10413253

#SPJ1

5 0

2 years ago

What is the approximation of pi to the nearest hundredth ?

saveliy_v [14]

Answer:

3.14

That is the value of pi to the second decimal point

7 0

3 years ago

Solve the equation with quadratic formula

Alla [95]

Answer:

x=4+9i,:x=4-9i

Step-by-step explanation:

put it into the quadratic formula

x=-(b+/- sqr b^2-4ac)/2a

8+/- sqr 64-97/2

7 0

3 years ago

Read 2 more answers

Rewrite the equation A = 1 2 ( b 1 + b 2 ) h A = 1 2 b 1 + b 2 h to show the expression that can be used to find the value of h.

Wittaler [7]

H=2A/b1+b2. Hope this helps!!

5 0

3 years ago

At time t=0 sec, a tank contains 15 oz of salt dissolved in 50 gallons of water. Then brine containing 88oz of salt per gallon o

maria [59]

Answer:

a) $s(t) =400- 385 e^{-\frac{1}{10} t}$

b) $s(t=25min) =400- 385 e^{-\frac{1}{10}25}=368.397$

Step-by-step explanation:

Part a

Assuming that the original concentration of salt is 8 oz/gal and that the rate of in is equal to the rate out = 5 gal/min.

For this case we know that the rate of change can be expressed on this way:

$Rate change = In-Out$

And we can name the rate of change as $\frac{ds}{dt}=rate change$

And our variable s would represent the amount of salt for any time t.

We know that the brine containing 8oz/gal and the rate in is equal to 5 gal/min and this value is equal to the rate out.

For the concentration out we can assume that is $\frac{s}{50gal}$

And now we can find the expression for the amount of salt after time t like this:

$\frac{dS}{dt}= 8 \frac{oz}{gal}(5\frac{gal}{min}) -\frac{s}{50gal} 5 \frac{gal}{min} =40\frac{oz}{min}- \frac{s}{10} \frac{oz}{min}$

And we have this differential equation:

$\frac{dS}{dt} +\frac{1}{10} s = 40$

With the initial conditions y(0)=15 oz

As we can see we have a linear differential equation so in order to solve it we need to find first the integrating factor given by:

$\mu = e^{\int \frac{1}{10} dt }= e^{\frac{1}{10} t}$

And then in order to solve the differential equation we need to multiply with the integrating factor like this:

$e^{\frac{1}{10} t} s = \int 40 e^{\frac{1}{10} t} dt$

$e^{\frac{1}{10} t} s = 400 e^{\frac{1}{10} t} +C$

Now we can divide both sides by $e^{\frac{1}{25} t}$ and we got:

$s(t) =400 + C e^{-\frac{1}{10} t}$

Now we can apply the initial condition in order to solve for the constant C like this:

$15 = 400+C$

$C=-385$

And then our function would be given by:

$s(t) =400- 385 e^{-\frac{1}{10} t}$

Part b

For this case we just need to replace t =25 and see what we got for the value of the concentration:

$s(t=25min) =400- 385 e^{-\frac{1}{10}25}=368.397$

7 0

3 years ago

Hello everyone (this is 6th grade math)​

Hello everyone (this is 6th grade math)