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Vika [28.1K]
2 years ago
12

A rectangle piece of paper has a width that is 3 inches less than it's length. It is cut in half along a diagonal to create two

congruent right triangles with areas of 44 square inches.Which statements are true
Mathematics
2 answers:
yarga [219]2 years ago
7 1

Answer:

A. The area of the rectangle is 88 square inches.

C. The equation x² – 3x – 88 = 0 can be used to solve for the length of the rectangle.

Step-by-step explanation:

The question is incomplete. Here is the complete question.

A rectangular piece of paper has a width that is 3 inches less than its length. It is cut in half along a diagonal to create two congruent right triangles with areas of 44 square inches. Which statements are true? Check all that apply.

A.The area of the rectangle is 88 square inches.

B.The equation x(x – 3) = 44 can be used to solve for the dimensions of the triangle.

C.The equation x2 – 3x – 88 = 0 can be used to solve for the length of the rectangle.

D.The triangle has a base of 11 inches and a height of 8 inches.

The rectangle has a width of 4 inches.

If a rectangular piece of paper has a width that is 3 inches less than it's length and it is cut in half along a diagonal to create two congruent right triangles with areas of 44 square inches, the width of the rectangle will be equal to the base of the each of the triangle, and the length equal to the height of the triangle.

Let the length of the rectangular piece of paper be L. If its width is 3 inches less than it's length, then W = L-3

The base of the triangle = L-3 and the height = L

Area of the triangle = 1/2 * base * height

Area of the triangle = 1/2 (L-3)L

If the area of the congruent triangles is 44in², then;

44 = 1/2 (L-3)L

88 = (L-3)L

88 = L²-3L

L²-3L-88=0

L²-11L+8L-88=0

L(L-11)+8(L-11)=0

(L-11)(L+8)= 0

L = 11in

This shows that the height of the triangle is 11 inches.

The base of the triangle = 11-3 = 8 inches

The area of the rectangle = Length*Breadth

= 11* 8

= 88in²

Based on the calculation above, the following statements are true.

-The area of the rectangle is 88 square inches.

-The equation x² – 3x – 88 = 0 can be used to solve for the length of the rectangle.

Guest1 year ago
1 0

It's A,C,D on EDGE

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One of the industrial robots designed by a leading producer of servomechanisms has four major components. Components’ reliabilit
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Answer:

a) Reliability of the Robot = 0.7876

b1) Component 1: 0.8034

    Component 2: 0.8270

    Component 3: 0.8349

    Component 4: 0.8664

b2) Component 4 should get the backup in order to achieve the highest reliability.

c) Component 4 should get the backup with a reliability of 0.92, to obtain the highest overall reliability i.e. 0.8681.

Step-by-step explanation:

<u>Component Reliabilities:</u>

Component 1 (R1) : 0.98

Component 2 (R2) : 0.95

Component 3 (R3) : 0.94

Component 4 (R4) : 0.90

a) Reliability of the robot can be calculated by considering the reliabilities of all the components which are used to design the robot.

Reliability of the Robot = R1 x R2 x R3 x R4

                                      = 0.98 x 0.95 x 0.94 x 0.90

Reliability of the Robot = 0.787626 ≅ 0.7876

b1) Since only one backup can be added at a time and the reliability of that backup component is the same as the original one, we will consider the backups of each of the components one by one:

<u>Reliability of the Robot with backup of component 1</u> can be computed by first finding out the chance of failure of the component along with its backup:

Chance of failure = 1 - reliability of component 1

                             = 1 - 0.98

                             = 0.02

Chance of failure of component 1 along with its backup = 0.02 x 0.02 = 0.0004

So, the reliability of component 1 and its backup (R1B) = 1 - 0.0004 = 0.9996

Reliability of the Robot = R1B x R2 x R3 x R4

                                         = 0.9996 x 0.95 x 0.94 x 0.90

Reliability of the Robot = 0.8034

<u>Similarly, to find out the reliability of component 2:</u>

Chance of failure of component 2 = 1 - 0.95 = 0.05

Chance of failure of component 2 and its backup = 0.05 x 0.05 = 0.0025

Reliability of component 2 and its backup (R2B) = 1 - 0.0025 = 0.9975

Reliability of the Robot = R1 x R2B x R3 x R4

                = 0.98 x 0.9975 x 0.94 x 0.90

Reliability of the Robot = 0.8270

<u>Reliability of the Robot with backup of component 3 can be computed as:</u>

Chance of failure of component 3 = 1 - 0.94 = 0.06

Chance of failure of component 3 and its backup = 0.06 x 0.06 = 0.0036

Reliability of component 3 and its backup (R3B) = 1 - 0.0036 = 0.9964

Reliability of the Robot = R1 x R2 x R3B x R4  

                = 0.98 x 0.95 x 0.9964 x 0.90

Reliability of the Robot = 0.8349

<u>Reliability of the Robot with backup of component 4 can be computed as:</u>

Chance of failure of component 4 = 1 - 0.90 = 0.10

Chance of failure of component 4 and its backup = 0.10 x 0.10 = 0.01

Reliability of component 4 and its backup (R4B) = 1 - 0.01 = 0.99

Reliability of the Robot = R1 x R2 x R3 x R4B

                                      = 0.98 x 0.95 x 0.94 x 0.99

Reliability of the Robot = 0.8664

b2) According to the calculated values, the <u>highest reliability can be achieved by adding a backup of component 4 with a value of 0.8664</u>. So, <u>Component 4 should get the backup in order to achieve the highest reliability.</u>

<u></u>

c) 0.92 reliability means the chance of failure = 1 - 0.92 = 0.08

We know the chances of failure of each of the individual components. The <u>chances of failure</u> of the components along with the backup can be computed as:

Component 1 = 0.02 x 0.08 = 0.0016

Component 2 = 0.05 x 0.08 = 0.0040

Component 3 = 0.06 x 0.08 = 0.0048

Component 4 =  0.10 x 0.08 = 0.0080

So, the <u>reliability for each of the component & its backup</u> is:

Component 1 (R1BB) = 1 - 0.0016 = 0.9984

Component 2 (R2BB) = 1 - 0.0040 = 0.9960

Component 3 (R3BB) = 1 - 0.0048 = 0.9952

Component 4 (R4BB) = 1 - 0.0080 = 0.9920

<u>The reliability of the robot with backups</u> for each of the components can be computed as:

Reliability with Component 1 Backup = R1BB x R2 x R3 x R4

                                                              = 0.9984 x 0.95 x 0.94 x 0.90

Reliability with Component 1 Backup = 0.8024

Reliability with Component 2 Backup = R1 x R2BB x R3 x R4

                                                              = 0.98 x 0.9960 x 0.94 x 0.90

Reliability with Component 2 Backup = 0.8258

Reliability with Component 3 Backup = R1 x R2 x R3BB x R4

                                                               = 0.98 x 0.95 x 0.9952 x 0.90

Reliability with Component 3 Backup = 0.8339

Reliability with Component 4 Backup = R1 x R2 x R3 x R4BB

                                                              = 0.98 x 0.95 x 0.94 x 0.9920

Reliability with Component 4 Backup = 0.8681

<u>Component 4 should get the backup with a reliability of 0.92, to obtain the highest overall reliability i.e. 0.8681. </u>

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