Been stuck on this question for 20 minutes, can someone please help.

Mathematics

1 answer:

Alexeev081 [22]3 years ago

6 0

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Find the solutions to 6x^2- 54x = 0

Nina [5.8K]

Answer:

A. x = 9

D. x = 0

Step-by-step explanation:

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$6x^{2} -54x=0$

$a=6\\b=-54\\c=0$

substitute in the formula

$x=\frac{-(-54)(+/-)\sqrt{-54^{2}-4(6)(0)}} {2(6)}$

$x=\frac{54(+/-)54} {12}$

$x=\frac{54(+)54} {12}=9$

$x=\frac{54(-)54} {12}=0$

therefore

The solutions are x=0 and x=9

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3 years ago

How many solutions can be found for the equation 4z + 2(z - 4) = 3z + 11

Gnom [1K]

4z+2(z-4)=3z+11 perform indicated multiplication on left side

4z+2z-8=3z+11 combine like terms on left side

6z/-8=3z+11 add 8 to both sides

6z=3z+19 subtract 3z from both sides

3z=19 divide both sides by 3

z=19/3

z=6 1/3

So there is only one solution when z=6 1/3

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3 years ago

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ira [324]

Answer:

Step-by-step explanation: