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postnew [5]
3 years ago
11

9/5 - (-2 7/10) and write it as a fraction and simplest form

Mathematics
2 answers:
bogdanovich [222]3 years ago
8 0

Answer:

4 1/2

Step-by-step explanation:

(-2 7/10) = (-2.7)

9/5 = 1.8

1.8 - (-2.7)

= 1.8 + 2.7

= 4.5 or 4 1/2

mars1129 [50]3 years ago
5 0

Answer:

probably 4.5 if im not mistaking

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vaieri [72.5K]
We know that In right angle triangle c^2 = a^2 + b^2.
19^2 = 3^2 + b^2
b = (361 - 9)^1/2 = 18.761663039294
Now for angles
 sin(A) = a/c = 3/19
A = sin{-1}(3/19) = 9.0847202873911
A = 9.0847202873911
sin(B) = b/c = 18.761663039294 / 19
B = sin{-1}(18.761663039294 / 19) = 80.915279712614
B = 80.915279712614

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3 years ago
I have 12 shirts, 4 pairs of shoes, 5 pants, and 3 watches. How many days can I go without wearing the same 4 items?
Nataly [62]
You can go 720 days without wearing the same 4 items. 
6 0
4 years ago
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La medida del vector se llama:​
Doss [256]

Step-by-step explanation:

suususisissiaisisjsjx

8 0
3 years ago
Calculation without distribution
Viefleur [7K]

Answer:

$2.69

Step-by-step explanation:

The prices of the 3 books are:

$2.50, $4.95, $6

The total is $2.50 + $4.95 + $6 = $13.45

The sale is 20% off.

20% of $13.45 =

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7 0
3 years ago
a total eclipse in 2003 lasted 2 17/24,minutes in 2005 it lasted 3/8 of a minute how much longer did it last in 2003
Doss [256]

Answer:

Total eclipse was 2\frac{1}{3} minutes longer than in 2005.

Step-by-step explanation:

It is given in the question that total eclipse in 2003 lasted 2\frac{17}{24} minutes and in 2005 it lasted 3/8 of a minute.

We have to calculate how much longer did it last in 2003.

So we have to subtract the duration in 2003 by duration in 2005

Total time taken in 2003- total time taken in 2005

=2\frac{17}{24}-\frac{3}{8}

\frac{48+17}{24}-\frac{3}{8}

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=\frac{65-9}{24}

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\frac{7}{3}

=2\frac{1}{3} minutes


3 0
4 years ago
Read 2 more answers
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