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katrin2010 [14]
3 years ago
8

Instead of the three dots, write a digit to make the fraction reducible. (Find all possible cases.) 77/... 333

Mathematics
2 answers:
dlinn [17]3 years ago
7 0

Answer:

129

Step-by-step explanation:

If you want it reduceable to a unit fraction, then the reciprocal would be equal to an integer. Make an equation, "(333+x)/77=y," and you see the positive side is just under 5, so you plug in "5 = y," and get 52 = x. That isn't a 3 digit number, so you plug in "6 = y," and get 129 = x.

yarga [219]3 years ago
7 0

Answer:

3 or 4

Step-by-step explanation:

77 has factors 7 and 11

To be reducible, the denominator must have one of these factors too

For the common factor to be 11:

77/3333

7/303

For the common factor to be 7:

77/4333

11/619

For the common factor to be 77:

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Mr. Roble is buying a lot of music on iTunes this week. On Monday, his bank account was at $30. On Tuesday, he bought 7 songs at
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Answer:

Step-by-step explanation:

7 songs at $1.99 each= 13.93

30-13.93= 16.07

16.07 + 10= 26.07

4 songs for $7.99= 31.96

26.07- 31.96= -5.89

so his pay check was $5.89

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3 years ago
What is 3(x-6)+(4x+12)-6x simplify.
Bezzdna [24]

Answer:

x-30

Step-by-step explanation:

Firstly, you do the first step, which is, 3x-18+4x+12-6x, then you continue simplifying it: 7x-18+12-6x.

Keep it on and on, the next step: x-18+12, add 18 and 12 together you get 30.

So the final answer is x-30

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3 years ago
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sladkih [1.3K]

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2 years ago
Which of the equations are true identities? a. (3x-2y)(2y-3x)=9x^2-4y^2 b. (2k+5r)(2k-5r)=4k^2-25r^2
krok68 [10]

Answer:

b.(2k+5r)(2k-5r)

Step-by-step explanation:

We have to find the equation which are true identity.

a.(3-2y)(2y-3x)

(3x-2y)(2y-3x)=3x(2y-3x)-2y(2y-3x)

=6xy-9x^2-4y^2+6xy

(3-2y)(2y-3x)\neq 9x^2-4y^2

b.(2k+5r)(2k-5r)

(2k)^2-(5r)^2

Using identity

(a+b)(a-b)=a^2-b^2

4k^2-25r^2

It is true.

Hence, option b is true.

7 0
3 years ago
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