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NeTakaya
3 years ago
8

1000:150:600 in lowest terms

Mathematics
2 answers:
Temka [501]3 years ago
7 0

Step-by-step explanation:

dividing every number by 10.

100:15:50

dividing every number by 5.

20:3:10.

this is the lowest term.

hope this helps you.

icang [17]3 years ago
5 0

Answer:

20:3:12

Step-by-step explanation:

divide first all of them by 10 it gives 100:15:60 then divide all of them by 5 it gives 20:3:12

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Please answer this question !! Thank you !! Will give brainliest !! Really important !!
Kobotan [32]

Answer:

0 = 3x -2y -6

Step-by-step explanation:

The general form of the equation of a line is 0 = Ax + By + C

y = 3/2x -3

Subtract y from each side

0 = 3/2x -y -3

Multiply each side by 2

0 = 3x -2y -6

8 0
3 years ago
The least common multiples of 4,10,12,15
lisabon 2012 [21]
Find the least common multiples of 4, 10, 12, 15

multiples of   4:   
4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80

multiples of 10: 10, 20, 30, 40, 50, 60, 70, 80, 90, 100, 110, 120

multiples of 12: 12, 24, 36, 48, 60, 72, 84, 96, 108, 120

multiples of 15: 15, 30, 45, 60, 75, 90, 105, 120, 135, 150

The least common multiple of 4, 10, 12, and 15 is 60.

4 x 15 = 60
10 x 6 = 60
12 x 5 = 60
15 x 4 = 60
6 0
3 years ago
Suppose that state’s record high
skelet666 [1.2K]
102 minus -14 = the range of 116 degrees.
6 0
4 years ago
Read 2 more answers
Which statement about sqrt x-5 minus sqrt x=5 true?
Allushta [10]

Answer:

9 is an extraneous solution.

Step-by-step explanation:

Extraneous solution is a solution that when plugged into the equation do not hold true.

Here we are given an algebraic equation in terms of single variable x as:

\sqrt{x-5}-\sqrt{x}=5-----(1)

Now, we will solve this equation to obtain the solution as:

on squaring both side of the equation we obtain:

(\sqrt{x-5}-\sqrt{x})^2=5^2

x-5+x-2\sqrt{x-5}\sqrt{x}=25\\\\2x-5-2\sqrt{x-5}\sqrt{x}=25\\\\2x-30=2\sqrt{x-5}\sqrt{x}\\\\on\ dividing\ both\ side\ by\ 2\ we\ obtain:\\\\x-15=\sqrt{x-5}\sqrt{x}

Again on squaring both side of the equation we obtain:

x^2+225-30x=(x-5)x\\\\\x^2+225-30x=x^2-5x\\\\225=-5x+30x\\\\225=25x\\\\x=9

Now when we plug x=9 back to the original equation i.e. equation (1) we get:

\sqrt{9-5}-\sqrt{9}=5\\\\\sqrt{4}-\sqrt{9}=5\\\\2-3=5\\\\-1=5

Hence, the equation does not hold true.

Hence, 9 is a extraneous solution

5 0
3 years ago
Perform the indicated operation.
rewona [7]

Answer:

your answer should be 7 1/24

6 0
4 years ago
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