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tiny-mole [99]
3 years ago
7

Answer this question please...

Mathematics
1 answer:
Gnesinka [82]3 years ago
5 0
No values of z make the equations true.
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What are the solutions to the quadratic equation (5y + 6)2 = 24? y = StartFraction negative 6 + 2 StartRoot 6 EndRoot Over 5 End
Paladinen [302]

Answer:

there are two solutions:

a) y=\frac{-6+2\sqrt{6} }{5}, and

b) y=\frac{-6-2\sqrt{6} }{5}

Step-by-step explanation:

In the equation: (5y+6)^2=24, since a perfect square with the unknown "y" is isolated on the left of the equal sign, we start by applying the square root on both sides of the equality, and then on isolating the unknown:

(5y+6)^2=24\\\sqrt{(5y+6)^2} =+/-\sqrt{24} \\(5y+6)=+/-\sqrt{6*4} \\(5y+6)=+/-2\sqrt{6}\\5y=-6+/-2\sqrt{6}\\y=\frac{-6+/-2\sqrt{6} }{5}

Therefore there are two solutions:

a) y=\frac{-6+2\sqrt{6} }{5}, and

b) y=\frac{-6-2\sqrt{6} }{5}

10 0
3 years ago
Read 2 more answers
The temperature in the morning is 18.6°C. By noon the temperature rises 8.5°C.
Harman [31]

Answer:

27.1°C

Step-by-step explanation:

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