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pickupchik [31]
3 years ago
8

Seth took 270 pictures for the yearbook And 180 for the school paper. Which proportion can be used to determine p,the percent of

the total numbers of pictures he took that were for the yearbook?
F: p/100=270/450. H: p/100= 450/270
G: p/100=180/450. J: p/100=270/180
Mathematics
2 answers:
sdas [7]3 years ago
4 0
I just had a test on this last week :)

F is correct

the total amount of pictures is 450 (270 + 180)
the part of 450 in the yearbook is 270
270/450 is one of the fractions
since it is a percent, p/100 is the other fraction
seraphim [82]3 years ago
3 0
F because, if you take 270/450=.6 or 60% and that would be your p, and 60%/100=.6

Enjoy!=)
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Please help?
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Step-by-step explanation:

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A_{\triangle IJK}=\dfrac{1}{2}\cdot IJ\cdot JK\cdot \sin 120^{\circ}=\dfrac{1}{2}\cdot 2\cdot 3\cdot \dfrac{\sqrt{3}}{2}=\dfrac{3\sqrt{3}}{2}\ un^2\\ \\ \\A_{\triangle KLM}=\dfrac{1}{2}\cdot KL\cdot LM\cdot \sin 120^{\circ}=\dfrac{1}{2}\cdot 8\cdot 2\cdot \dfrac{\sqrt{3}}{2}=4\sqrt{3}\ un^2\\ \\ \\A_{\triangle MNI}=\dfrac{1}{2}\cdot MN\cdot NI\cdot \sin 120^{\circ}=\dfrac{1}{2}\cdot 3\cdot 8\cdot \dfrac{\sqrt{3}}{2}=6\sqrt{3}\ un^2\\ \\ \\

2. Note that

A_{\triangle IJK}=A_{\triangle IAK}=\dfrac{3\sqrt{3}}{2}\ un^2 \\ \\ \\A_{\triangle KLM}=A_{\triangle KAM}=4\sqrt{3}\ un^2 \\ \\ \\A_{\triangle MNI}=A_{\triangle MAI}=6\sqrt{3}\ un^2

3. The area of hexagon IJKLMN is the sum of the area of all triangles:

A_{IJKLMN}=2\cdot \left(\dfrac{3\sqrt{3}}{2}+4\sqrt{3}+6\sqrt{3}\right)=23\sqrt{3}\ un^2

Another way to solve is to find the area of triangle KIM be Heorn's fomula, where all sides KI, KM and IM can be calculated using cosine theorem.

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