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Charra [1.4K]
4 years ago
7

Find the least common denominator for these two rational expressions. 3/2n -2/5n

Mathematics
2 answers:
Zielflug [23.3K]4 years ago
8 0
Every rational expression can be written in infinitely many equvialent  forms which is 3/3n-15/2.
zmey [24]4 years ago
3 0
For 2n and 5n the least common denominator will be 10n
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Given: O is the midpoint of line MN<br> OM=OW<br> Prove: OW=ON
grigory [225]
Given:
O is the midpoint of line MN
OM = OW

To prove:  OW = ON

<u>Statement</u>                                 <u>Reason</u>
1> OM = OW  -------------------------> Given
2> OM = ON ---------------------------> O is the midpoint of line MN
                                                           i.e Point O bisects line MN
3> OM = OW --------------------------> From statement <1>
4> ON = OW  -------------------------> OM = ON (Statement <2>)
     OW = ON
              
                                                                    <u>proved!!</u>
   


6 0
4 years ago
Three boxes are shipped on a truck,. Each box has a base of 16 square feet. Two of the boxes have a height of 3 feet and one box
vodka [1.7K]
I believe the answer is 176 cubic feet
7 0
3 years ago
Read 2 more answers
a dish has 21,245,673 bacteria in it. The population of bacteria will decline by 80% every day. How many bacteria will be presen
Ivahew [28]

Answer:

33993.0768

Step-by-step explanation:

You can multiply the number of bacteria by (0.2^4). The 0.2 is for the 20% that  left after each day. The (^4) is for the 4 days that have passed.

6 0
3 years ago
What is the bottom answers?
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4 0
2 years ago
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FIND THE SUM OF 20 TERMS OF THE ARITHMETIC SERIES IN WHICH 3rd TERM IS 7 AND 7th TERM IS 2 MORE THAN THREE TIMES ITS 3rd TERM
algol13

Answer:

740

Step-by-step explanation:

The n th term of an arithmetic series is

a_{n} = a₁ + (n - 1)d

where a₁ is the first term and d the common difference

Given a₃ = 7 and a₇ = (3 × 7) + 2 = 21 + 2 = 23 , then

a₁ + 2d = 7 → (1)

a₁ + 6d = 23 → (2)

Subtract (1) from (2) term by term

4d = 16 ( divide both sides by 4 )

d = 4

Substitute d = 4 into (1)

a₁ + 2(4) = 7

a₁ + 8 = 7 ( subtract 8 from both sides )

a₁ = - 1

The sum to n terms of an arithmetic series is

S_{n} = \frac{n}{2} [ 2a₁ + (n - 1)d ] , thus

S_{20} = \frac{20}{2} [ (2 × - 1) + (19 × 4) ]

     = 10(- 2 + 76) = 10 × 74 = 740

8 0
3 years ago
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