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3 years ago

11

Jenny is cooking a chicken a chicken needs to be cooked 40 mins per kilogram plus extra 20 mins how much will it take to cook a

chicken weighing 3 kilograms

1 answer:

Nostrana [21]3 years ago

7 0

If it is 3kg, it will be:
40mins x 3 = 120mins
Add the extra 20 mins, it will be 140 mins. This equates to two hours and 20 minutes.

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Answer:

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Step-by-step explanation:

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Neporo4naja [7]

To answer this item, we let x be the distance that Jimmy drove from his origin and back. Given the speeds, the time it takes for him to travel to and fro are,
t1 = x / 46
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Step-by-step explanation:

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3 years ago

I need help with my algebra 2. I also want to know how to solve this.

Lisa [10]

Add 1 to both sides:

$\sqrt{x+3} = x+1$

In cases like this, we have to remember that a root is always positive, so we can square both sides only assuming that

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But since we can only accept solutions greater than -1, we discard $x=-2$ and accept $x=1$ .

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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}
$

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3 years ago