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aivan3 [116]
2 years ago
10

Would someone help me 2/3 2/1 8/9 7/9 breast to least​

Mathematics
2 answers:
Tema [17]2 years ago
4 0

Answer:

First of all, it's the Greatest to least! Second of all, 2/1, 8/9, 7/9, and 2/3.

Step-by-step explanation:

2/1 is 2, 8/9 is 1 more ninth than 7/9, and 2/3 is less third than 1.

Salsk061 [2.6K]2 years ago
3 0

Answer:

It is 2/1 then 2/3 then 7/8 then 8/9  i hope this helps you friend

Step-by-step explanation:

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bogdanovich [222]

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3 0
1 year ago
20 points
gavmur [86]

Answer:

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6 0
2 years ago
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Find the amount financed and the finance charge in dollars for the following problem. Choose the right answers. Joshua Bennett p
Tpy6a [65]
25.83 x 6 = 154.98

If the purchased television is 245.95 and you make 6 payments that total in
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40.25% = 0.4025 

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98.99 dollars off the original price.

245.95 - 98.99 = 146.96

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4 0
2 years ago
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Jeffery babysits for 4 dollars per hour, he also works as a math tutor for 7 dollars per hour. He is only allowed to work 13 hou
Oksana_A [137]

Given:

Jeffery babysits for 4 dollars per hour and works as a math tutor for 7 dollars per hour.

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He wants to make at least 65 dollars.

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The system of inequalities to represent this situation.

Solution:

Let x be the number of hours he babysits per week and y be the number of hours h works as a math tutor.

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x+y\leq 13

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x\geq 0, y\geq 0.

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4 0
3 years ago
Can you answer these please many thanks
givi [52]

Given: (a) 2(x+3)=2x+6 and (b) 4(y-3)=4y-12.

To find: The expanded form of the given expressions.

(c) 4(m+n)=4m+4n       (using a(b+c)=ab+ac)

(d) 3(5-q)=15-3q           (using a(b-c)=ab-ac)

(e) 5(2c+1)=10c+5          (using a(b+c)=ab+ac)

(f) 3(2x-5)=6x-15          (using a(b-c)=ab-ac)

(g) 7(4b-1)=28b-7          (using a(b-c)=ab-ac)

(h) 3(2x+y-5)=3((2x+y)-5)

⇒3(2x+y-5)=3(2x+y)-15         (using a(b-c)=ab-ac)

⇒3(2x+y-5)=6x+3y-15            (using a(b+c)=ab+ac)

(i) 2(6a-4b+3)=2((6a-4b)+3)

⇒2(6a-4b+3)=2(6a-4b)+6         (using a(b+c)=ab+ac)

⇒2(6a-4b+3)=12a-8b+6            (using a(b-c)=ab-ac)

(j)6(m+n+p)=6((m+n)+p)

⇒ 6(m+n+p)=6(m+n)+6p           (using a(b+c)=ab+ac)

⇒ 6(m+n+p)=6m+6n+6p            (using a(b+c)=ab+ac)

(k) y(y+2)=y^2+2y          (using a(b+c)=ab+ac)

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(m) n(4-n)=4n-n^2        (using a(b-c)=ab-ac)

(n) a(b+c)=ab+ac           (using a(b+c)=ab+ac)

(o) s(3s-4)=3s^2-4s        (using a(b-c)=ab-ac)

(p) 2x(x+5)=2x^2+10x     (using a(b+c)=ab+ac)

(q) 4y(x-3)=4xy-12y      (using a(b-c)=ab-ac)

(r) 5a(2b-5)=10ab-25a     (using a(b-c)=ab-ac)

(s) 4a(3b+2c)=12ab+8ac    (using a(b+c)=ab+ac)

(t) 5p(4p-5q)=20p^2-25pq    (using a(b-c)=ab-ac)

6 0
3 years ago
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