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AysviL [449]
3 years ago
9

4 students where is scheduled to give book reports in 1 hour. After the first report 2/3 hour remaind. The next two reports took

1/6 hour and 1/4 hour. What fraction of the hour remained?
Mathematics
2 answers:
kaheart [24]3 years ago
6 0
First we need to work out how long each report took to write.
First report= 1/3 hour
Second report= 1/6 hour
Third report= 1/4 hour
Then we need to find multiple of 3,4 and 6. Which is 12 because all of these numbers can make 12.
First report= 4/12
Second report= 2/12
Third report=3/12
If we add them all together this makes 9/12.
To find what fraction of an hour is left we take 9/12 away from 12/12.
This equals 3/12 hour.
If your specifically asked to give it in the simplest form you need to find a common factor that can divide the whole fraction down. In this fraction you can choose the number 3. If you divide this down.
Your answer is 1/4 of an hour left.
Julli [10]3 years ago
5 0
When adding fractions you need a common denominator so lets look at what "fractions" we are dealing with:2/3 1/6 and 1/4.

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NO LINKS!!! Part 2: Find the Lateral Area, Total Surface Area, and Volume. Round your answer to two decimal places.​
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Answer:

<h3><u>Question 7</u></h3>

<u>Lateral Surface Area</u>

The bases of a triangular prism are the triangles.

Therefore, the Lateral Surface Area (L.A.) is the total surface area excluding the areas of the triangles (bases).

\implies \sf L.A.=2(10 \times 6)+(3 \times 6)=138\:\:m^2

<u>Total Surface Area</u>

Area of the isosceles triangle:

\implies \sf A=\dfrac{1}{2}\times base \times height=\dfrac{1}{2}\cdot3 \cdot \sqrt{10^2-1.5^2}=\dfrac{3\sqrt{391}}{4}\:m^2

Total surface area:

\implies \sf T.A.=2\:bases+L.A.=2\left(\dfrac{3\sqrt{391}}{4}\right)+138=167.66\:\:m^2\:(2\:d.p.)

<u>Volume</u>

\sf \implies Vol.=area\:of\:base \times height=\left(\dfrac{3\sqrt{391}}{4}\right) \times 6=88.98\:\:m^3\:(2\:d.p.)

<h3><u>Question 8</u></h3>

<u>Lateral Surface Area</u>

The bases of a hexagonal prism are the pentagons.

Therefore, the Lateral Surface Area (L.A.) is the total surface area excluding the areas of the pentagons (bases).

\implies \sf L.A.=5(5 \times 6)=150\:\:cm^2

<u>Total Surface Area</u>

Area of a pentagon:

\sf A=\dfrac{1}{4}\sqrt{5(5+2\sqrt{5})}a^2

where a is the side length.

Therefore:

\implies \sf A=\dfrac{1}{4}\sqrt{5(5+2\sqrt{5})}\cdot 5^2=43.01\:\:cm^2\:(2\:d.p.)

Total surface area:

\sf \implies T.A.=2\:bases+L.A.=2(43.01)+150=236.02\:\:cm^2\:(2\:d.p.)

<u>Volume</u>

\sf \implies Vol.=area\:of\:base \times height=43.011193... \times 6=258.07\:\:cm^3\:(2\:d.p.)

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