A rancher has 200 ft of fencing to enclose two adjacent rectangular corrals.

Ill give you Brainlyest! I need help ASAP work shown, please!!<br> problems 8-15

8_murik_8 [283]

Using the segment addition theorem, the solutions are:

5. x = 17

6. x = 7

7. x = 14

BC = 27

CD = 61

BD = 88

8. AB = 26

9. LJ = 46

10. x = 3

11. FG = 15

12. QS = 34

13. BC = 26

14. EG = 19

15. QS = 68

<h3>How to Apply the Segment Addition Theorem?</h3>

The segment addition theorem would be used to solve the problems as shown below:

5. UW = UV + VW [segment addition theorem]

Substitute

6x - 35 = 19 + 4x - 20

6x - 4x = 35 + 19 - 20

2x = 34

x = 17

6. HJ = HI + IJ [segment addition theorem]

Substitute

7x - 27 = 3x - 5 + x - 1

7x - 3x - x = 27 - 5 - 1

3x = 21

x = 7

7. BD = BC + CD [segment addition theorem]

Substitute

7x - 10 = 4x - 29 + 5x - 9

7x - 4x - 5x = 10 - 29 - 9

-2x = -28

x = 14

BC = 4x - 29 = 4(14) - 29 = 27

CD = 5x - 9 = 5(14) - 9 = 61

BD = 7x - 10 = 7(14) - 10 = 88

8. BC = BD

Substitute

2x + 1 = 5x - 26

2x - 5x = -1 - 26

-3x = -27

x = 9

AB = 43 - BC

AB = 43 - 2x + 1 = 43 - 2(9) + 1 = 26

9. 7x - 10 = 9x - 11 - (x + 3)

7x - 10 = 9x - 11 - x - 3

7x - 9x + x = 10 - 11 - 3

-x = -4

x = 4

LJ = 28 + 7x - 10 = 28 + 7(4) - 10

LJ = 46

10. 8x + 11 = 12x - 1

8x - 12x = -11 - 1

-4x = -12

x = 3

11. 11x - 7 = 3x + 9

11x - 3x = 7 + 9

8x = 16

x = 2

FG = 11x - 7 = 11(2) - 7

FG = 15

12. 5x - 3 = 21 - x

5x + x = 21 + 3

6x = 24

x = 4

QS = 2(5x - 3) = 10x - 6 = 10(4) - 6

QS = 34

13. 8x - 20 = 2(3x - 1)

8x - 20 = 6x - 2

8x - 6x = 20 - 2

2x = 18

x = 9

BC = AB = 3x - 1 = 3(9) - 1

BC = 26

14. 5x - 1 = 7x - 13

5x - 7x = 1 - 13

-2x = -12

x = 6

EG = 6x - 4 - 13 = 6(6) - 4 - 13

EG = 19

15. RT - ST = RS

Substitute

8x - 43 - (4x - 1) = 2x - 4

8x - 43 - 4x + 1 = 2x - 4

4x - 42 = 2x - 4

4x - 2x = 42 - 4

2x = 38

x = 19

QS = 2(RS)

QS = 2(2x - 4) = 4x - 8 = 4(19) - 8

QS = 68

Learn more about the segment addition theorem on:

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8 0

2 years ago

Graph g(x) = -3(x + 2)2 + 2. Compare the graph to the graph of f(x) = x2. Give me five points that you use to graph g(x). Then d

PolarNik [594]

The answer would be -20 because you add and subtract your numbers to get 20

4 0

2 years ago

The volume of a rectangular pyramid is 900 cubic centimeters.It has a length of 10 centimeters and a width of 9 centimeters. Wha

MariettaO [177]

The height of a rectangular pyramid is 30 cm.

volume of rectangular pyramid = (length x width x height) / 3

Given:
Volume = 900 cm³
Length = 10 cm
Width = 9 cm
Height = ?

V = lwh/3
900 cm³ = (10cm * 9cm * h)/3
900 cm³ * 3 = (90cm² * h)/3 * 3
2700 cm³ = 90cm² * h
2700 cm³/90cm² = 90cm² * h/90 cm²
30 cm = h

to check:
900 cm³ = (10 cm * 9 cm * 30 cm)/3
900 cm³ = 2700 cm³/3
900 cm³ = 900 cm³

3 0

3 years ago

A student rolls a die 15 times and records the results. This is the first set of data. The same student rolls a different die 15

Stells [14]

Answer:

C

Step-by-step explanation:

unimodel: A unimodal distribution is a distribution with one clear peak or most frequent value. The values increase at first, rising to a single peak where they then decrease.

3 0

3 years ago

Read 2 more answers

Mrs. Backstrom built a small wooden box for her cat. The box has a length of one and one tenth feet, a width of 1.6 feet, and a

likoan [24]

Answer:

$\text{Volume of cat box}=3.52\text{ ft}^3$

Step-by-step explanation:

We have been given that Mrs. Backstrom built a small wooden box for her cat. The box has a length of one and one tenth feet, a width of 1.6 feet, and a height of 2 feet. We are asked to find the volume of the box.

Let us convert one and one tenth feet into decimal.

$1\frac{1}{10}=\frac{11}{10}=1.1$

Since the cat box is in form of cuboid, so its volume will be equal to the product of length, width and height as:

$\text{Volume of cat box}=1.10\text{ ft}\times 1.6\text{ ft}\times 2\text{ ft}$

$\text{Volume of cat box}=3.52\text{ ft}^3$

Therefore, the volume of the cat box is 3.53 cubic feet.

8 0

3 years ago