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Leya [2.2K]
3 years ago
7

A student solved the following problem and made an error: Triangles ABC and DEF. Angles A and F are congruent and measure 135 de

grees. Coordinates for the vertices are at A 0, 2 and B 2, 4 and C 0, 0 and D 2, 0 and E 4, 4 and F 4, 2. Line 1 Segment AC equals 2. Segment FE equals 2. Segment AC is congruent to segment FE Line 2 ∠A ≅ ∠F Line 3 Length of segment AB. A (0, 2) B (2, 4) d equals square root of quantity x sub 2 minus x sub 1 squared plus quantity y sub 2 minus y sub 1 squared, d equals square root of quantity 0 minus 2 all squared plus quantity 2 minus 4 all squared, d equals square root of negative 2 squared plus negative 2 squared, d equals square root of 4 plus 4, d equals square root of 8 segment AB = 2.83 Line 4 Length of segment DE. D (2, 0) E (4, 4) d equals square root of quantity x sub 2 minus x sub 1 squared plus quantity y sub 2 minus y sub squared, d equals square root of quantity 2 minus 4 all squared plus quantity 0 minus 4 all squared, d equals square root of negative 2 squared plus negative 4 squared, d equals square root of 4 plus 16, then d equals square root of 20 segment DE = 4.47 Line 5 segment AB is congruent with segment DE Line 6 triangle ABC is congruent with triangle FDEby SAS In which line did the student make the first mistake? Line 2 Line 5 Line 3 Line 4
Mathematics
1 answer:
nignag [31]3 years ago
7 0

The first false statement in the proof as it stands is in Line 5, where it is claimed that a line of length 2.83 is congruent to a line of length 4.47. This mistake cannot be corrected by adding lines to the proof.

_____

The first erroneous tactical move is in Line 4, where the length of DE is computed, rather than the length of FD. This mistake can be corrected by adding lines to the proof.

A correct SAS proof would use segment FD in Line 4, so it could be argued that the first mistake is there.

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Step-by-step explanation:

90 + 122 + 4x + 19 + 9x - 8 + 7x - 3 = 540

220 + 20x = 540

-220             -220

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(4*-2)-(-2*8)=
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Maryann's friends loved her bulletin board , and 4 of them asked her to make one for their homes. She offered to make each of he
bulgar [2K]

Complete Question:

To create a bulletin board, Maryann is planning to stretch fabric from burlap sacks over cardstock. She collects 6 kg of burlap sacks. The area density of the burlap is 0.4 kg/m².

Maryann's friends loved her bulletin board , and 4 of them asked her to make one for their homes. She offered to make each of her friends a replica of her bulletin board that measures 2 meters by 3 meters . How many kilograms of burlap should Maryann collect to complete the bulletin boards for her friends?

Answer:

To complete the bulletin boards for her friends, Maryanne should collect 6 kg of burlap more

Step-by-step explanation:

Mass of burlap sacks that Maryann collects = 6 kg

Area density of burlap = 0.4 kg/m²

Area Density = Mass / Area........(1)

The Area of bulletin board that can be made with the 6 kg of burlap sand can be calculated by using equation (1)

0.4 = 6 / Area

Area = 6/0.4

Area of bulletin board that can be made from 6 kg= 15 m² of bulletin board  

The area of bulletin board that Maryann made for herself = 2 * 3 = 6m²

To make similar bulletin boards for her four friends, the total area of bulletin board made for her 4 friends = 6 * 4 = 24m²

To calculate the mass of burlap required to make a 24m² bulletin board, use equation (1)

0.4 = Mass / 24

Mass = 0.4 * 24

Mass = 9.6 kg

The mass of burlap used to make Maryann's bulletin board alone:

0.4 = Mass/6

Mass = 2.4 kg

Out of the 6kg that she got, she has used 2.4 kg for herself

The remaining mass of burlap = 6 - 2.4 = 3.6 kg

Therefore, to complete the bulletin board for her friend, she should collect (9.6 - 3.6) = 6 kg of bulletin more

5 0
3 years ago
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