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yanalaym [24]
3 years ago
6

Find the perimeter of the image below: Figure QRSTU is shown. Q is at 2, 0. R is at 4, 5. S is at 8, 7. T is at 6, 4. U is at 10

, 3. 25.8 units 26.1units 27.5 units 28.6 units

Mathematics
2 answers:
rewona [7]3 years ago
4 0
The correct answer should be B which is 26.1 

The full answer is 26.16 
Svet_ta [14]3 years ago
4 0

Answer:

Hence, the perimeter of the figure is:

(√29+ 2√5 +√13+ √17+√73) units=26.1 units.

Step-by-step explanation:

The perimeter of the figure is the length of all the line segments.

i.e. Line segment QR,RS,ST,TU and QU.

We know that the distance between two points (a,b) and (c,d) is given as:

\sqrt{(c-a)^2+(d-b)^2}.

  • Length of segment Q(2,0)R(4,5) is equal to the distance between these the coordinates of these vertex.

i.e. distance between (2,0) and (4,5) is:

\sqrt{(4-2)^2+(5-0)^2}\\\\=\sqrt{2^2+5^2}\\\\=\sqrt{4+25}\\\\=\sqrt{29}

Hence length of line segment QR is: √29 units.

  • Length of segment R(4,5)S(8,7) is equal to the distance between these the coordinates of these vertex.

i.e. the distance between (4,5) and (8,7) is:

\sqrt{(8-4)^2+(7-5)^2}\\\\=\sqrt{4^2+2^2}\\\\=\sqrt{16+4}\\\\=\sqrt{20}=2\sqrt{5}

Hence length of line segment RS is: 2√5 units.

  • Length of segment S(8,7)T(6,4) is equal to the distance between these the coordinates of these vertex.

i.e. the distance between (8,7) and (6,4) is:

\sqrt{(8-6)^2+(7-4)^2}\\\\=\sqrt{2^2+3^2}\\\\=\sqrt{4+9}\\\\=\sqrt{13}

Hence length of line segment RS is: √13 units.

  • Length of segment T(6,4)U(10,3) is equal to the distance between these the coordinates of these vertex.

i.e. the distance between (6,4) and (10,3) is:

\sqrt{(10-6)^2+(3-4)^2}\\\\=\sqrt{4^2+(-1)^2}\\\\=\sqrt{16+1}\\\\=\sqrt{17}

Hence length of line segment RS is: √17 units.

  • Length of segment U(10,3)Q(2,0) is equal to the distance between these the coordinates of these vertex.

i.e. the distance between (10,3) and (2,0) is:

\sqrt{(10-2)^2+(3-0)^2}\\\\=\sqrt{8^2+3^2}\\\\=\sqrt{64+9}\\\\=\sqrt{73}

Hence length of line segment RS is: √73 units.

Hence, the perimeter of the figure is:

(√29+ 2√5 +√13+ √17+√73) units=26.1 units

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The key to solving this one is to work it backwards.

-- They had 19 peaches left at the end of the day.

-- During the final hour, the 1/5 of a peach that they sold left them with 19,
so they had 19-1/5 before they sold the 1/5 of a peach.
The 19-1/5 was 4/5 of what they had at the beginning of the final hour.
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-- During the 3rd hour, the 3/4 of a peach that they sold left them with 24,
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Do the same for the 2nd hour.

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Answer:

P(4≤x≤7) = 2/3

Step-by-step explanation:

We'll begin by obtaining the sample space (S) i.e possible outcome of rolling both dice at the same time. This is illustrated below:

1,1 1,2 1,3 1,4 1,5 1,6

2,1 2,2 2,3 2,4 2,5 2,6

3,1 3,2 3,3 3,4 3,5 3,6

4,1 4,2 4,3 4,4 4,5 4,6

5,1 5,2 5,3 5,4 5,5 5,6

6,1 6,2 6,3 6,4 6,5 6,6

Adding the outcome together, the sample space (S) becomes:

2 3 4 5 6 7

3 4 5 6 7 8

4 5 6 7 8 9

5 6 7 8 9 10

6 7 8 9 10 11

7 8 9 10 11 12

Next, we shall obtain the event of 4≤x≤7. This is illustrated below:

4 5 6 7

4 5 6 7

4 5 6 7

4 5 6 7

4 5 6 7

4 5 6 7

Finally, we shall determine P(4≤x≤7). This can be obtained as follow:

Element in the sample space, n(S) = 36

Element in 4≤x≤7, n(4≤x≤7) = 24

Probability of 4≤x≤7, P(4≤x≤7) = ?

P(4≤x≤7) = n(4≤x≤7) / nS

P(4≤x≤7) = 24/36

P(4≤x≤7) = 2/3

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