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Leno4ka [110]
2 years ago
5

The boxed definition of absolute value states that |a|=-a if a is a negative number. Explain why |a| is always nonnegative, even

though |a|=-a for negative values of a.
Mathematics
1 answer:
Temka [501]2 years ago
7 0

Final answer:

Since absolute values determine the distance between the number and the value whether the value is positive or negative. As distance is always positive.

Thus, |a| is always nonnegative, even though |a|=-a for negative values of a.

Step-by-step explanation:

Step 1

It is said that |a| is always nonnegative even though even though |a|=-a for negative values of a.

Step 2

This is because by the definition of an absolute value, any real number inside an absolute value symbol || will always be positive.

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castortr0y [4]

Answer:

221.4m^{3}

Step-by-step explanation:

To work out the volume of this shape you will first need to work out the area of the triangle. To work out the area of the triangle you would multiply the base of 12 by the height of 4.1, which gives you 49.2. You would then need to divide 49.2 by 2, which gives you 24.6. This is because the formula for the area of a triangle is the same as that of a square, however a triangle is half of a square with the same length and width. To work out the volume you would multiply the area of the triangle, which is 24.6 by the height of 9, which gives you 221.4m^{3}

1) Multiply 12 by 4.1.

12*4.1=49.2

2) Divide 49.2 by 2.

49.2/2=24.6

3) Multiply 24.6 by 9.

24.6*9=221.4m^{3}

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3 years ago
The elevation of driskill mountain Louisiana is 163 meters above sea level death Valley has an elevation of -86 meters right any
timama [110]

Answer:

-86

Step-by-step explanation:

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8 0
3 years ago
Read 2 more answers
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natita [175]
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I decided to included the steps just in case you needed them

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Describe the translation of the function from the parent function f(x) = x².
kkurt [141]
\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\\\

\begin{array}{rllll} 
% left side templates
f(x)=&{{  A}}({{  B}}x+{{  C}})+{{  D}}
\\ \quad \\
y=&{{  A}}({{  B}}x+{{  C}})+{{  D}}
\\ \quad \\
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\\ \quad \\
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\\ \quad \\
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\end{array}

\bf \begin{array}{llll}
% right side info
\bullet \textit{ stretches or shrinks horizontally by  } {{  A}}\cdot {{  B}}\\\\
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\\\\

\end{array}\\

\bf \begin{array}{llll}
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\bullet \textit{ vertical shift by }{{  D}}\\
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\bullet \textit{ period of }\frac{2\pi }{{{  B}}}
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now, notice the template above... now let's see your function \bf y=x^2+4\implies 
\begin{array}{llllll}
y=&1(&1x&+&0)^2+&4\\
&A&B&&C&D
\end{array}

so.. what do you think was the shift then?
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80\geqp-20
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