What needs to be corrected in the following construction for copying∠ABC with point D as the vertex?

2 answers:

5 0

I think you forgot to attach the diagram along with the question. I am answering the question based on my research. "The third arc should cross the second arc" that needs to be corrected <span>in the following construction for copying∠ABC with point D as the vertex. The correct option among all the options that are given in the question is option "c".</span>

seropon [69]3 years ago

5 0

The third arc should cross the second arc.

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What is the value of x?

Eduardwww [97]

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asambeis [7]

$\bf \qquad \qquad \textit{double proportional variation}\\\\
\begin{array}{llll}
\textit{\underline{y} varies directly with \underline{x}}\\
\textit{and inversely with \underline{z}}
\end{array}\implies y=\cfrac{kx}{z}\impliedby 
\begin{array}{llll}
k=constant\ of\\
\qquad variation
\end{array}\\\\
-------------------------------\\\\$

$\bf \begin{cases}
p=pedalisi\\
w=weight\\
s=\textit{sitting height}
\end{cases}\quad 
\begin{array}{llll}
%pelidisi, varies directly as the cube root of a person's weight in grams and inversely as the person's sitting height in centimeters.
\textit{pelidisi varies directly}\\
\textit{as cube root of weight}\\
\textit{and inversely to }\\
\textit{sitting height}
\end{array}\implies p=\cfrac{k\sqrt[3]{w}}{s}\\\\
-------------------------------$

$\bf \textit{we know that }
\begin{cases}
w=48,820\\
s=78.7\\
p=100
\end{cases}\implies 100=\cfrac{k\sqrt[3]{48820}}{78.7}
\\\\\\
100\cdot 78.7=k\sqrt[3]{48820}\implies \cfrac{7870}{\sqrt[3]{48820}}=k
\\\\\\
thus\qquad \boxed{p=\cfrac{\frac{7870}{\sqrt[3]{48820}}\sqrt[3]{w}}{s}}
\\\\\\
\textit{now, what is \underline{p} when }
\begin{cases}
w=54,688\\
s=72.6
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3 years ago

beaus recipefor granola bars calls for 3 1/2 cups of oatmeal. he only have 1/6 cup scoop.how many scoops of oatmeal will he need

Stells [14]

<h2>Answer:</h2>

The number of scoops of oatmeal he will need for the recipe is:

21 scoops.

<h2>Step-by-step explanation:</h2>

It is given that:

Beaus recipe for granola bars calls for 3 1/2 cups of oatmeal.

i.e. the cups of oatmeal required is:

$3\dfrac{1}{2}\ cups=\dfrac{3\times 2+1}{2}\ cups$

i.e.

$\dfrac{7}{2}\ \text{cups\ of\ oatmeal\ is\ required}$

Also, he has:

$\dfrac{1}{6}\ \text{cup scoop}$

This means that:

$\text{Number of scoops of oatmeal he need}=\dfrac{\text{Number of cups of oatmeal}}{\text{Number of cup scoop he has}}\\\\i.e.\\\\\text{Number of scoops of oatmeal he need}=\dfrac{\dfrac{7}{2}}{\dfrac{1}{6}}\\\\i.e.\\\\\text{Number of scoops of oatmeal he need}=\dfrac{7\times 6}{1\times 2}\\\\i.e.\\\\\text{Number of scoops of oatmeal he need}=21\ \text{scoops}$