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Arisa [49]
3 years ago
11

Want brainliest pick a number 1-3 first to answer gets it

Mathematics
1 answer:
Alecsey [184]3 years ago
8 0

Answer:

3

Step-by-step explanation:

3 because 3 is a good number....lol

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Answer with workings please
icang [17]
AB = 9 cm
BC = 6cm

CD = 7 cm
AE = 6 cm

3BC = AB
3ED = AE

  AB = AE
  BC    ED
    ⁹/₃ = ⁶/ₓ
3 · 6 = 9 · x
   18 = 9x
    9      9
     2 = x

ED = 2 cm
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3 years ago
Problem-solving Practice
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First, what's 3 to the power of 8?
Then what's 3 to the power of 10

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Read 2 more answers
Select from the drop-down menus to correctly complete each statement.
tino4ka555 [31]

Answer:

Right of -7 and left of -1

Step-by-step explanation:

-7 is in the left and -4 is in the right

-7<4

-4 in the left and -1 in the right

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4 0
3 years ago
Solve.
Nikitich [7]
The answer is C: 10\text{ }^1/_2. Here are the details:

\text{Equation:}\\ 6\text{ }^1/_3+10\text{ }^1/_2\stackrel{?}{=}26\text{ }^1/_6\\ \\ \text{Start with the integers.}\\ 6+10=16\stackrel{?}{=}26\text{ }^1/_6\\ \\ \text{...then with the fractions, but rewrite them first to make it easier.}\\ ^1/_3+\text{ }^1/_2\stackrel{\text{rewrite}}{\to}\text{ }^2/_6+\text{ }^3/_6=\text{ }^5/_6\stackrel{?}{=}26\text{ }^1/_6\\&#10;\\&#10;\text{Add 'em up!}\\&#10;16\text{ }^5/_6

\text{Equation:}\\&#10;16\text{ }^5/_6+3\text{ }^5/_6\stackrel{?}{=}26\text{ }^1/_6\\&#10;\\&#10;\text{Start with the integers.}\\&#10;16+3=19\stackrel{?}{=}26\text{ }^1/_6\\&#10;\\&#10;\text{...then with the fractions, but rewrite them first to make it easier.}\\&#10;^5/_6+\text{ }^5/_6=\text{ }^{10}/_6\stackrel{\text{rewrite}}{\to}\text{ }^5/_3\stackrel{\text{rewrite}}{\to}1\text{ }^2/_3\stackrel{?}{=}26\text{ }^1/_6\\&#10;\\&#10;\text{Add 'em up!}\\&#10;20\text{ }^2/_3

\text{Last equation:}\\ 20\text{ }^2/_3+5\text{ }^1/_2\stackrel{?}{=}26\text{ }^1/_6\\ \\ \text{Start with the integers.}\\ 20+5=25\stackrel{?}{=}26\text{ }^1/_6\\ \\ \text{...then with the fractions, but rewrite them first to make it easier.}\\ ^2/_3+\text{ }^1/_2\stackrel{\text{rewrite}}{\to}\text{ }^4/_6+\text{ }^3/_6=\text{ }^7/_6\stackrel{\text{rewrite}}{\to}1\text{ }^1/_6\stackrel{?}{=}26\text{ }^1/_6\\&#10;\\&#10;\text{Add the integer and fraction together.}\\&#10;25+1\text{ }^1/_6\stackrel{?}{=}26\text{ }^1/_6

26\text{ }^1/_6\stackrel{\checkmark}{=}26\text{ }^1/_6
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3 years ago
Consider two bases B = {b1, b2, b3} and C = {c1, c2, c3} for a vector space V such that
Helen [10]
I’m a failure if you can just say “yes you are” in the comments
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3 years ago
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