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Trava [24]
3 years ago
13

How do you solve question 43 and 44?

Mathematics
1 answer:
Sauron [17]3 years ago
5 0
<1=90=3y-6
3y=96
y=96:3
y=32
9x=6z=90
x=10
z=90:6
z=15

44. BD=AC
     2x-1+3y+5=4x-y+1
     2x+3y+4=4x-y+1
      3y+y+4-1=4x-2x
       4y+3=2x
       3y+5=2x-1
        3y+5=(4y+3)-1
         3y+5=4y+2
          5-2=4y-3y
          y=3
2x=4y+3=4*3+3=15
   x=15:2
x=7,5
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2) Ordered pairs are (20,20) ,(16,17), (12,14), (8,11)

3) Ordered pairs are     (1,40),(3,22), (7,13), (15,7.5)

Step-by-step explanation:

1) Rule 1: Add 2 starting from 9

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Sequence 1       Sequence 2       Ordered Pair

9                           5                           (9,5)

11                          8                           (11,8)

13                          11                           (13,11)

15                          14                           (15,14)

2) Rule 1: Subtract 4 starting from 20

Rule 2: Subtract 3 starting from 20

Sequence 1 is generated using Rule 1 so, it starts with 20 and we subtract 4 to get the next number.

Sequence 2 is generated using Rule 2 so, it starts with 20 and we subtract 3 to get the next number.

Sequence 1       Sequence 2       Ordered Pair

20                         20                        (20,20)

16                          17                         (16,17)

12                          14                         (12,14)

8                            11                         (8,11)

3) Rule 1: Multiply by 2 then add 1 starting from 1

Rule 2: Divide by 2 then add 4 starting from 40

Sequence 1 is generated using Rule 1 so, it starts with 1 and then we multiply by 2 and then add 2 to get the next number.

Sequence 2 is generated using Rule 2 so, it starts with 40 and then we multiply by 2 and add 2 to get the next number.

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1                           40                            (1,40)

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15                          7.5                           (15,7.5)

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