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Neko [114]
3 years ago
14

Subtract 9y - 3 from 6y +1

Mathematics
2 answers:
Jlenok [28]3 years ago
4 0
9y - 3 - 6y + 1
9y - 6y - 3 + 1
3y - 2
Alexus [3.1K]3 years ago
3 0

Answer: 3y +2

Step-by-step explanation:

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Solve the following quadratic function
Leya [2.2K]

Answer:

x = ± 10

Step-by-step explanation:

to solve the equation let y = 0 , that is

100 - x² = 0 ( add x² to both sides )

100 = x² ( take square root of both sides )

± \sqrt{100} = x , that is

x = ± 10

8 0
2 years ago
Match each expression in the left column with the correct product in the right column.
lesya692 [45]

Answer:

-6

-6i

6i

6

Step-by-step explanation:

1) √4 . √-3 . √-3

$ \sqrt{4} = 2 $

$ \sqrt{-3} . \sqrt{-3} = (\sqrt{-3})^2 $

$ \sqrt{4} . (\sqrt{-3})^2 = 2 \times -3 = $ -6

2) √-4 . √-3 . √-3

$ \sqrt{-4} = 2i $ .

Therefore, $ \sqrt{-4} . \sqrt{-3} . \sqrt{-3} = 2. \sqrt{-1}  \times -3 = 2i \times (-3) =  $ - 6i

3) √4 . √3 . √-3

$ \sqrt{4} = 2 $

$ \sqrt{3} . \sqrt{-3} = (\sqrt{3})^2 . \sqrt{-1} $

$ \implies 2 \times 3i = $ 6i

4) √4 . √3 . √3

$ \sqrt{4} = 2 $

$ \sqrt{3} . \sqrt{3} = (\sqrt{3})^2  = 3 $

Therefore, √4 . √3 . √3 = 2 . 3 = 6

5 0
3 years ago
Select the graph for the solution of the open sentence. Click until the correct graph appears. |x| > 1
Lisa [10]

Answer:

See attachment

Step-by-step explanation:

The given inequality is |x|>1

By the definition of the absolute value function, we obtain the compound inequality.

-x>1\:or\:x>1

When we simplify we get:

x1

The graph for the solution set is shown in the attachment

7 0
3 years ago
Math geometry question 3, Thanks if u help!
Studentka2010 [4]

Answer:

7 and 1

Step-by-step explanation:

√50 is the sum of the square of two legs

The answer is likely to be 7 and 1 (because 7^2 + 1^1 = 50)

7 0
3 years ago
The figures below are similar. The labeled sides are corresponding.
aev [14]

Answer: 24 cm

Step-by-step explanation:

P_{square} = side *4

It is multiplied by 4 because a square has 4 equal sides and 4 equal angles.

P = 6 * 4 = 24

6 0
2 years ago
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