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Ne4ueva [31]
2 years ago
12

What is Q1 AND Q3 AND IQR OF THE NUMBERS2,12,52,33,8,14

Mathematics
2 answers:
Sedaia [141]2 years ago
7 0
It is 25.

Q1 is 8
Q3 is 33
Wittaler [7]2 years ago
4 0
25 is the answer.

Q1=8
Q3=33

---------
I did the work.
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WHAT IS X³-27 SIMPLIFIED
Eduardwww [97]

Answer:

<u>It</u><u> </u><u>is</u><u> </u><u>(</u><u>x</u><u> </u><u>-</u><u> </u><u>3</u><u>)</u><u>³</u><u> </u><u>-</u><u> </u><u>9</u><u>x</u><u>(</u><u>3</u><u> </u><u>-</u><u> </u><u>x</u><u>)</u>

Step-by-step explanation:

Express 27 in terms of cubes, 27 = 3³:

=  {x}^{3}  -  {3}^{3}

From trinomial expansion:

{(x - y)}^{3}  = (x - y)(x - y)(x - y) \\

open first two brackets to get a quadratic equation:

{(x - y)}^{3}  = ( {x}^{2}  - 2xy +  {y}^{2} )(x - y)

expand further:

{(x - y)}^{3}  =  {x}^{3}   - y {x}^{2}  - 2y {x}^{2}  + 2x {y}^{2}  + x {y}^{2}  -  {y}^{3}  \\  {(x - y)}^{3}  =  {x}^{3}  -  {y}^{3}  + 3x {y}^{2}  - 3y {x}^{2}  \\  {(x - y)}^{3}  =  {x}^{3}  -  {y}^{3}  + 3xy(y - x) \\  \\ { \boxed{( {x}^{3} -  {y}^{3} ) =  {(x - y)}^{3}   - 3xy(y - x)}}

take y to be 3, then substitute:

( {x}^{3}  - 3^3) =  {(x - 3)}^{3}  - 9x(3 - x)

5 0
2 years ago
HELPPPP Write the equation of a line in slope intercept form with a slope of -3/4 and a y-intercept of -9.
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Answer:

y=-3/4x-9

Step-by-step explanation:

plz give brainliest!! hope this helps!

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3 years ago
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jeyben [28]

Answer:

6

Step-by-step explanation:

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Whole Numbers include both positive and negative numbers. t or f
Alex

Answer:

t

Step-by-step explanation:

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3 years ago
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Write an algebraic expression to match the graph<br><br> please help!
skelet666 [1.2K]

the graphed equation is:

y = -|x| + 1

<h3></h3><h3>How to find the algebraic expression that matches the graph?</h3>

On the graph, we can see an absolute value equation, which opens downwards, and passes through the points (1, 0) and (0,1).

The general absolute value equation is:

y = a*|x - b| + c

Because the vertex is at the point (0, 1), we know that:

b = 0, c = 1

Replacing that we get:

y = a*|x| + 1

Now, this must be zero when we evaluate in x = 1, then:

0 = a*|1| + 1

a = -1

Then the graphed equation is:

y = -|x| + 1

If you want to learn more about absolute value functions:

brainly.com/question/3381225

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