What is the greatest fraction

Mathematics

2 answers:

fiasKO [112]3 years ago

7 0

2/3 is the greatest fraction

ioda3 years ago

3 0

2/3=16/24
5/8=15/24
the greatest fraction is 2/3

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Add 4/9 +2/9 and simplify answer

dimulka [17.4K]

Answer: 6/9 = 2/3

Step-by-step explanation: The answer is 2/3 or 6/9 is because you add 2 to 4 and that is 6 you keep the denominator the same because both fractions have the same denominator and you get 6/9, 6/9 is simplified to 2/3

7 0

3 years ago

1) Solve by using the perfect squares method. x2 + 8x + 16 = 0

Bond [772]

1)
$x^2+8x+16=0 \\
(x+4)^2=0 \\
x+4=0 \\
\boxed{x=-4}$

2)
$x^2-5x-6=0 \\
x^2-6x+x-6=0 \\
x(x-6)+1(x-6)=0 \\
(x+1)(x-6)=0 \\
x+1=0 \ \lor \ x-6=0 \\
x=-1 \ \lor \ x=6 \\
\boxed{x=-1 \hbox{ or } x=6}$

3)
$\hbox{a perfect square:} \\ (x-a)^2=x^2-2xa+a^2 \\ \\ 2xa=20x \\ a=\frac{20x}{2x} \\ a=10 \\ \\ a^2=10^2=100 \\ \\ \hbox{the expression:} \\ x^2-20x+100 \\ \\ \boxed{\hbox{100 should be added to the expression}}$

4)
$x^2+8x-8=0 \\ \\
a=1 \\ b=8 \\ c=-8 \\ \Delta=b^2-4ac=8^2-4 \times 1 \times (-8)=64+32=96 \\
\sqrt{\Delta}=\sqrt{96}=\sqrt{16 \times6}=4\sqrt{6} \\ \\
x=\frac{-b \pm \sqrt{\Delta}}{2a}=\frac{-8 \pm 4\sqrt{6}}{2 \times 1}=\frac{2(-4 \pm 2\sqrt{6})}{2}=-4 \pm 2\sqrt{6} \\
\boxed{x=-4-2\sqrt{6} \hbox{ or } x=-4+2\sqrt{6}}$

5)
$2x^2+12x=0 \\
2x(x+6)=0 \\
2x=0 \ \lor \ x+6=0 \\
x=0 \ \lor \ x=-6 \\
\boxed{x=-6 \hbox{ or } x=0}$

6)
$2x^2-2x-1=0 \\ \\
a=2 \\ b=-2 \\ c=-1 \\ \Delta=b^2-4ac=(-2)^2-4 \times 2 \times (-1)=4+8=12 \\
\sqrt{\Delta}=\sqrt{12}=\sqrt{4 \times 3}=2\sqrt{3} \\ \\
x=\frac{-b \pm \sqrt{\Delta}}{2a}=\frac{-(-2) \pm 2\sqrt{3}}{2 \times 2}=\frac{2 \pm 2\sqrt{3}}{2 \times 2}=\frac{2(1 \pm \sqrt{3})}{2 \times 2}=\frac{1 \pm \sqrt{3}}{2} \\
\boxed{x=\frac{1-\sqrt{3}}{2} \hbox{ or } x=\frac{1+\sqrt{3}}{2}}$

7)
$x^2-x+2=0 \\ \\
a=1 \\ b=-1 \\ c=2 \\
\Delta=b^2-4ac=(-1)^2-4 \times 1 \times 2=1-8=-7 \\ \\
\boxed{\hbox{the discriminant } \Delta=-7}$

8)
$3x^2-6x+1=0 \\ \\
a=3 \\ b=-6 \\ c=1 \\ \Delta=b^2-4ac=(-6)^2-4 \times 3 \times 1=36-12=24 \\ \\
\boxed{\hbox{the discriminant } \Delta=24} \\ \\
\hbox{if } \Delta\ \textless \ 0 \hbox{ then there are no real roots} \\
\hbox{if } \Delta=0 \hbox{ then there's one real root} \\
\hbox{if } \Delta\ \textgreater \ 0 \hbox{ then there are two real roots} \\ \\
\Delta=24\ \textgreater \ 0 \\
\boxed{\hbox{the equation has two real roots}}$

9)
$y=2x^2+x-3 \\ \\ a=2 \\ b=1 \\ c=-3 \\ \Delta=b^2-4ac=1^2-4 \times 2 \times (-3)=1+24=25 \\ \\ \hbox{the function has two zeros} \\ \boxed{\hbox{the parabola has 2 points in common with the x-axis}} \\ \\ a\ \textgreater \ 0 \hbox{ so the parabola ope} \hbox{ns upwards} \\ \boxed{\hbox{the vertex lies below the x-axis}}$

10)
$y=x^2-12x+12 \\ \\
a=1 \\ b=-12 \\ c=12 \\ \Delta=b^2-4ac=(-12)^2-4 \times 1 \times 12=144-48=96 \\ \\ \hbox{the function has two zeros} \\ \boxed{\hbox{the parabola has 2 points in common with the x-axis}} \\ \\ a\ \textgreater \ 0 \hbox{ so the parabola ope} \hbox{ns upwards} \\ \boxed{\hbox{the vertex lies below the x-axis}}$

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