Round 141 decimal point 999 to the nearest tenth hundredth ten and hundred

1 answer:

3 0

141.999 rounded to the nearest hundredth AND tenth is 142 Rounded to the nearest ten is 140 Rounded to the nearest hundred is 100 If this answer was helpful to you, please mark it as brainliest! I would appreciate that very much!

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<img src="https://tex.z-dn.net/?f=%20%7C7%20%2B%208x%7C%20%20%3E%205" id="TexFormula1" title=" |7 + 8x| > 5" alt=" |7 + 8x|

ExtremeBDS [4]

The absolute value is defined as

$|x|=\begin{cases}x&\text{if }x\ge0\\-x&\text{if }x$

So for example, if x = 3, then |x| = |3| = 3, since 3 is positive. On the other hand, if x = -5, then |x| = |-5| = -(-5) = 5, since -5 is negative. The absolute value is always positive.

For the inequality |7 + 8x| > 5, you consider the two cases where the argument to the absolute value (the expression you find inside the bars) is either positive or negative.

• If 7 + 8x ≥ 0, then |7 + 8x| = 7 + 8x, so that

$|7+8x|>5\implies 7+8x>5 \implies 8x>-2 \implies x>-\dfrac14$

• Otherwise, if 7 + 8x < 0, then |7 + 8x| = -(7 + 8x), so that

$|7+8x|>5\implies-(7+8x)>5\implies7+8x$

The solution to the inequality is the union of these two intervals.

3 0

3 years ago

HELP HOW AM I MEANT TO ANSWER THIS QUESTION I WOULD ASO LIKE ANSWER. THX

NemiM [27]

Answer: the smallest is 2^5, 6^2, 3^4 and the largest is 5^3

5 0

2 years ago

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a 3 pound pork loin can be cut into 10 pork chops of equal weight. how many ounces are in each pork chop? please show your work

Artyom0805 [142]

Attached a solution and showed work.

3 0

3 years ago

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Y = (1/4)x^2 - (1/2)lnx..over the interval (1, 7e) ...what is the arc length ?

xenn [34]

So, f[x] = 1/4x^2 - 1/2Ln(x)
thus f'[x] = 1/4*2x - 1/2*(1/x) = x/2 - 1/2x 
thus f'[x]^2 = (x^2)/4 - 2*(x/2)*(1/2x) + 1/(4x^2) = (x^2)/4 - 1/2 + 1/(4x^2) 
thus f'[x]^2 + 1 = (x^2)/4 + 1/2 + 1/(4x^2) = (x/2 + 1/2x)^2 
thus Sqrt[...] = (x/2 + 1/2x)

6 0

3 years ago

Find the exact value of tan (arcsin (two fifths)). For full credit, explain your reasoning.

Hitman42 [59]

$\bf sin^{-1}(some\ value)=\theta \impliedby \textit{this simply means}
\\\\\\
sin(\theta )=some\ value\qquad \textit{now, also bear in mind that}
\\\\\\
sin(\theta)=\cfrac{opposite}{hypotenuse}\qquad 
\qquad 
% tangent
tan(\theta)=\cfrac{opposite}{adjacent}\\\\
-------------------------------\\\\$

$\bf sin^{-1}\left( \frac{2}{5} \right)=\theta \impliedby \textit{this simply means that}
\\\\\\
sin(\theta )=\cfrac{2}{5}\cfrac{\leftarrow opposite}{\leftarrow hypotenuse}\qquad \textit{now let's find the adjacent side}
\\\\\\
\textit{using the pythagorean theorem}\\\\
c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}$

$\bf \pm \sqrt{5^2-2^2}=a\implies \pm\sqrt{21}=a
\\\\\\
\textit{we don't know if it's +/-, so we'll assume is the + one}\quad \sqrt{21}=a\\\\
-------------------------------\\\\
tan(\theta)=\cfrac{opposite}{adjacent}\qquad \qquad tan(\theta)=\cfrac{2}{\sqrt{21}}
\\\\\\
\textit{and now, let's rationalize the denominator}
\\\\\\
\cfrac{2}{\sqrt{21}}\cdot \cfrac{\sqrt{21}}{\sqrt{21}}\implies \cfrac{2\sqrt{21}}{(\sqrt{21})^2}\implies \cfrac{2\sqrt{21}}{21}
$

7 0

2 years ago