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Elina [12.6K]
4 years ago
8

(a - b)(a 2 + ab + b 2) HELP

Mathematics
2 answers:
Blizzard [7]4 years ago
8 0
Use FOIL. 

(a - b)(a^2 + ab + b^2) 
a^3 + a^2b + b^2a - a^2b - b^2a - b^3
a^3 - b^3 <----------Answer 



Mila [183]4 years ago
7 0
Hey there :)

( a - b )( a² + ab + b² )
Let us distribute ( a - b ) into the other

a ( a² ) + a ( ab ) + a ( b² ) - b ( a² ) - b ( ab ) - b ( b² )
a³ + a²b + ab² - a²b - ab² - b³

We can further simplify
a²b - ab² = 0
ab² - ab² = 0

a³ - b³

You can see in the picture which I have attached. It is the 8th formula

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Answer:

Equation 1:  r = <u>4</u> +( <u>3</u> * <u>cos theta</u>  )

Equation 2: r = sqrt (<u> 5²</u> * <u>sin(2 theta)</u> )

Step-by-step explanation:

GRAPH 1:  

The first graph is a dimpled limacon.

General equation for dimpled limacon:

r = a + b cos theta                     ∴ if dimple is along the x- axis

r = a + b sin theta                      ∴ if dimple is along the y-axis

y-intercept : { a, -a }  = { 4, -4 }   ∴ the points at which limacon intersects y-axis

Negative side of x-axis = ( a – b ) ⇒ 1

Positive side of x-axis = ( a + b ) ⇒ 7

Subtract the value of a from sum of a and b to find b:

b = 7 – 4 ⇒ 3

Equation1:  r = <u>4</u> +( <u>3</u> * <u>cos theta</u>  )

GRAPH 2:  

The second graph is a lemniscates.  

General equation for lemniscates is:

r² = a² cos(2theta)                        ∴ if petals of graph are on coordinate axis

r² = a² sin(2 theta)                ∴ if petals of graph are not on coordinate axis

now, according to the graph:

a = 5 ⇒ a² = 25

angle of graph:  cos2θ, simply divide 360° by 2:

\frac{360}{2} ⇒ 180°

The petals cannot be on coordinate axis, we start from 45° and then the next petal will be on:

45° + 180° = 225°

Since the graph is not on the coordinate axis, so

r² = 5² sin(2 theta)   ⇒    r = sqrt ( 5² * sin(2 theta) )      

Equation 2: r = sqrt (<u> 5²</u> * <u>sin(2 theta)</u> )      

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Answer:

ARE ≅ ΔARC

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Step-by-step explanation:

We are given that in the two triangles ARE and ARC.

Sides AE ≅ AC and RE ≅ RC

Moreover, they have AR as their common side.

<em>So, we get that, in ΔARE and ΔARC, all the three corresponding sides are equal.</em>

Thus, by the SSS Congruence Postulate, ΔARE ≅ ΔARC.

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Step-by-step explanation:

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