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3 years ago

Rewrite the numbers from least to greatest 0.33,-0.12,-0.127,0.2,45/10

Mathematics

2 answers:

Sonbull [250]3 years ago

8 0

-0.127
-0.12
0.20
0.33
45/10
I think that is right

Minchanka [31]3 years ago

5 0

-0.127, -0.12, 0,2, 0.33, 45/10

This probably is a little more complicated than the average least to greatest problem. The part that used to throw me off was the fact that they all had different amounts of numbers behind the decimal. I learned to just add as many zeros BEHIND the given decimal to even it out.

Another thing that might be throwing you off is the negatives. I learned this pretty quickly, but I do know some people you didn't. The greater the number, the further left the order / number line it belongs. So, if it were -0.9 and -0.7 then -0.9 would have a lesser value, in this case. ONLY WITH NEGATIVES!!

Hope this helped!

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Answer:

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Step-by-step explanation:

1st boat:

Parabola equation:

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The x-coordinate of the vertex:

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Solve:

$c=10+a\\ \\80a+10+a=1\\ \\81a=-9\\ \\a=-\dfrac{1}{9}\\ \\b=-2a=\dfrac{2}{9}\\ \\c=10-\dfrac{1}{9}=\dfrac{89}{9}$

Parabola equation:

$y=-\dfrac{1}{9}x^2 +\dfrac{2}{9}x+\dfrac{89}{9}$

2nd boat:

Parabola equation:

$y=ax^2 +bx+c$

The x-coordinate of the vertex:

$x_v=-\dfrac{b}{2a}\Rightarrow -\dfrac{b}{2a}=0\\ \\b=0$

Equation:

$y=ax^2+c$

The y-coordinate of the vertex:

$y_v=a\cdot 0^2+c\Rightarrow c=-7$

Parabola passes through the point (-8,1), so

$1=a\cdot (-8)^2-7\\ \\64a-7=1$

Solve:

$a=-\dfrac{1}{8}\\ \\b=0\\ \\c=-7$

Parabola equation:

$y=\dfrac{1}{8}x^2 -7$

System of two equations:

$\left\{\begin{array}{l}y=-\dfrac{1}{9}x^2 +\dfrac{2}{9}x+\dfrac{89}{9}\\ \\y=\dfrac{1}{8}x^2 -7\end{array}\right.$

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