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nekit [7.7K]
3 years ago
8

What is the solution of -8/2y-8=5/y+4-7y+8/y^2-16 y = –4 y = –2 y = 4 y = 6

Mathematics
2 answers:
maria [59]3 years ago
4 0
(-8)/(2y-8)=(5/(y+4))-7y+(8/(y^2-16))
(-4)/(y-4)=(5/(y+4))-7y+(8/(y+4)(y-4))
((-4)(y+4))/((y+4)(y-4))=((5(y-4))/(y+4)(y-4))-(7y(y+4)(y-4))/(y+4)(y-4))+(8/(y+4)(y-4))
(-4(y+4))=(5(y-4))-(7y(y+4)(y-4))+8
-4y-16=5y-20-(7y(y^2-16))+8
-4y-16=5y-20-7y^3+112y+8
-4y-16=117y-7y^3-12
-4=(121-7y^2)(y)
None of these choices would be equal to -4

Fittoniya [83]3 years ago
4 0

Answer:

The answer is <em><u>y=6</u></em>!

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X + Y is equal to 5 x minus Y is equal to 4 find the value of find the value of x square minus y square ​
Grace [21]

Answer: 20

Step-by-step explanation:

x+y=5

x-y=4

---------

x+y=5

x=4+y

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(4+y)+y=5

4+y+y=5

 4+2y=5

      2y=1

        y=0.5

        x=4.5

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x²-y²=(4.5)²-(0.5)²=20.25-0.25=20

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3 years ago
2 Select the correct answer. Consider functions f and g. Which statement gives the best approximation of the solutions of the eq
kaheart [24]
  • Answer: are c and d

Step-by-step explanation: I think not sure.

6 0
2 years ago
Convert 72°F to °C (to the nearest tenth) using one of the following formulas: F = 9/5xC + 32 or C = (F - 32) x 5/9.
Fudgin [204]
First set up the equation:
C=(F-32)\times\frac{5}{9}
Then substitute in known values:
C=(72-32)\times\frac{5}{9}
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C=40\times\frac{5}{9}
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8 0
3 years ago
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PLS HELP ME ASAP FOR 9) A and B (SHOW WORK!!) + LOTS OF POINTS!
statuscvo [17]
A.. 9.5*3.6=34.2 then 34.2/5.7= 6 so 6 is your answer

B. 16.2*5.7= 92.34 then 92.34/3.6= 25.65
4 0
3 years ago
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Suppose brine containing 0.2 kg of salt per liter runs into a tank initially filled with 500 L of water containing 5 kg of salt.
Oliga [24]

Answer:

(a) 0.288 kg/liter

(b) 0.061408 kg/liter

Step-by-step explanation:

(a) The mass of salt entering the tank per minute, x = 0.2 kg/L × 5 L/minute = 1 kg/minute

The mass of salt exiting the tank per minute = 5 × (5 + x)/500

The increase per minute, Δ/dt, in the mass of salt in the tank is given as follows;

Δ/dt = x - 5 × (5 + x)/500

The increase, in mass, Δ, after an increase in time, dt, is therefore;

Δ = (x - 5 × (5 + x)/500)·dt

Integrating with a graphing calculator, with limits 0, 10, gives;

Δ = (99·x - 5)/10

Substituting x = 1 gives

(99 × 1 - 5)/10 = 9.4 kg

The concentration of the salt and water in the tank after 10 minutes = (Initial mass of salt in the tank + Increase in the mass of the salt in the tank)/(Volume of the tank)

∴ The concentration of the salt and water in the tank after 10 minutes =  (5 + 9.4)/500 = (14.4)/500 = 0.288

The concentration of the salt and water in the tank after 10 minutes = 0.288 kg/liter

(b) With the added leak, we now have;

Δ/dt = x - 6 × (14.4 + x)/500

Δ = x - 6 × (14.4 + x)/500·dt

Integrating with a graphing calculator, with limits 0, 20, gives;

Δ = 19.76·x -3.456 = 16.304

Where x = 1

The increase in mass after an increase in = 16.304 kg

The total mass = 16.304 + 14.4 = 30.704 kg

The concentration of the salt in the tank then becomes;

Concentration = 30.704/500 = 0.061408 kg/liter.

6 0
3 years ago
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