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3 years ago

Please please please help

Mathematics

2 answers:

tia_tia [17]3 years ago

3 0

It’s false!!! Hope this helped;)

yarga [219]3 years ago

3 0

Answer:

False

Step-by-step explanation:

4x+4+5y=9xy+4

Step 1: Add -9xy to both sides.

4x+5y+4+−9xy=9xy+4+−9xy

−9xy+4x+5y+4=4

Step 2: Add -5y to both sides.

−9xy+4x+5y+4+−5y=4+−5y

−9xy+4x+4=−5y+4

Step 3: Add -4 to both sides.

−9xy+4x+4+−4=−5y+4+−4

−9xy+4x=−5y

Step 4: Factor out variable x.

x(−9y+4)=−5y

Step 5: Divide both sides by -9y+4.

x(−9y+4)

−9y+4

−5y

−9y+4

−5y

−9y+4

Answer:

−5y

−9y+4

this does not add up as in um.. i forgot the word- but yea its false

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shtirl [24]

Answer:

y=8x

Step-by-step explanation:

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3 years ago

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Three consecutive odd integers are such that the square of the third integer is 1515 greater than greater than the sum of the sq

Ksju [112]

Let
x-----------> first <span>odd integer
x+2--------> second consecutive odd integer
x+4-------> third consecutive odd integer

we know that
(x+4)</span>²=15+x²+(x+2)²-------> x²+8x+16=15+x²+x²+4x+4
x²+8x+16=19+2x²+4x-------> x²-4x+3
x²-4x+3=0

using a graph tool----------> to calculate the quadratic equation
see the attached figure
the solution is
x=1
x=3

the answer is
the first odd integer x is 1
the second consecutive odd integer x+2 is 3
the third consecutive odd integer x+4 is 5

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3 years ago

A tank with a capacity of 1600 L is full of a mixture of water and chlorine with a concentration of 0.0125 g of chlorine per lit

Veronika [31]

Answer:

y(t) = 20 [1600^(-5/3)] x (1600-24t)^ (5/3)

Step-by-step explanation:

1) Identify the problem

This is a differential equation problem

On this case the amount of liquid in the tank at time t is 1600−24t. (When the process begin, t=0 ) The reason of this is because the liquid is entering at 16 litres per second and leaving at 40 litres per second.

2) Define notation

y = amount of chlorine in the tank at time t,

Based on this definition, the concentration of chlorine at time t is y/(1600−24t) g/ L.

Since liquid is leaving the tank at 40L/s, the rate at which chlorine is leaving at time t is 40y/(1600−24t) (g/s).

For this we can find the differential equation

dy/dt = - (40 y)/ (1600 -24 t)

The equation above is a separable Differential equation. For this case the initial condition is y(0)=(1600L )(0.0125 gr/L) = 20 gr

3) Solve the differential equation

We can rewrite the differential equation like this:

dy/40y = - (dt)/ (1600-24t)

And integrating on both sides we have:

(1/40) ln |y| = (1/24) ln (|1600-24t|) + C

Multiplying both sides by 40

ln |y| = (40/24) ln (|1600 -24t|) + C

And simplifying

ln |y| = (5/3) ln (|1600 -24t|) + C

Then exponentiating both sides:

e^ [ln |y|]= e^ [(5/3) ln (|1600-24t|) + C]

with e^c = C , we have this:

y(t) = C (1600-24t)^ (5/3)

4) Use the initial condition to find C

Since y(0) = 20 gr

20 = C (1600 -24x0)^ (5/3)

Solving for C we got

C = 20 / [1600^(5/3)] = 20 [1600^(-5/3)]

Finally the amount of chlorine in the tank as a function of time, would be given by this formula:

y(t) = 20 [1600^(-5/3)] x (1600-24t)^ (5/3)

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