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Tema [17]
3 years ago
9

What is the justification for the first step in proving the formula for factoring the sum of cubes?

Mathematics
1 answer:
Anna11 [10]3 years ago
5 0

Explanation:

The formula isnt correctly written, it should state:

a^3+b^3 = (a+b)(a^2-ab+b^2)

You have to start from (a+b)(a^2-ab+b^2)  and end in a³+b³. On your first step, you need to use the distributive property.

(a+b)(a^2-ab+b^2) = a*(a^2-ab+b^2) + b*(a^2-ab+b^2)

This is equal to

a*a^2-a*(ab) + a*b^2 + b*a^2-b*(ab) + b*b^2 = a^3 - a^2b + ab^2 +ba^2 -b^2a +b^3

Note that the second term, -a²b, is cancelled by the fourth term, ba², and the third term, ab², is cancelled by the fifht term, -b²a. Therefore, the final result is a³+b³, as we wanted to.

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Which of the following is the equation for the graph shown?
Allushta [10]
It would probably be the second option.

3 0
3 years ago
a magazine contains fourteen pages. You open to a random page, the page number is three or seven..... where do i start i need to
Tems11 [23]
This is a question of randomness so there really is no answer. I would guess 7 since that is the center page and you're more likely to open it up in or near the middle
7 0
3 years ago
Read 2 more answers
For two events A and B show that P (A∩B) ≥ P (A)+P (B)−1.
nordsb [41]

Answer:

<h3>For two events A and B show that P (A∩B) ≥ P (A)+P (B)−1.</h3>

By De morgan's law

(A\cap B)^{c}=A^{c}\cup B^{c}\\\\P((A\cap B)^{c})=P(A^{c}\cup B^{c})\leq P(A^{c})+P(B^{c}) \\\\1-P(A\cap B)\leq  P(A^{c})+P(B^{c}) \\\\1-P(A\cap B)\leq  1-P(A)+1-P(B)\\\\-P(A\cap B)\leq  1-P(A)-P(B)\\\\P(A\cap B)\geq P(A)+P(B)-1

which is Bonferroni’s inequality

<h3>Result 1: P (Ac) = 1 − P(A)</h3>

Proof

If S is universal set then

A\cup A^{c}=S\\\\P(A\cup A^{c})=P(S)\\\\P(A)+P(A^{c})=1\\\\P(A^{c})=1-P(A)

<h3>Result 2 : For any two events A and B, P (A∪B) = P (A)+P (B)−P (A∩B) and P(A) ≥ P(B)</h3>

Proof:

If S is a universal set then:

A\cup(B\cap A^{c})=(A\cup B) \cap (A\cup A^{c})\\=(A\cup B) \cap S\\A\cup(B\cap A^{c})=(A\cup B)

Which show A∪B can be expressed as union of two disjoint sets.

If A and (B∩Ac) are two disjoint sets then

P(A\cup B) =P(A) + P(B\cap A^{c})---(1)\\

B can be  expressed as:

B=B\cap(A\cup A^{c})\\

If B is intersection of two disjoint sets then

P(B)=P(B\cap(A)+P(B\cup A^{c})\\P(B\cup A^{c}=P(B)-P(B\cap A)

Then (1) becomes

P(A\cup B) =P(A) +P(B)-P(A\cap B)\\

<h3>Result 3: For any two events A and B, P(A) = P(A ∩ B) + P (A ∩ Bc)</h3>

Proof:

If A and B are two disjoint sets then

A=A\cap(B\cup B^{c})\\A=(A\cap B) \cup (A\cap B^{c})\\P(A)=P(A\cap B) + P(A\cap B^{c})\\

<h3>Result 4: If B ⊂ A, then A∩B = B. Therefore P (A)−P (B) = P (A ∩ Bc) </h3>

Proof:

If B is subset of A then all elements of B lie in A so A ∩ B =B

A =(A \cap B)\cup (A\cap B^{c}) = B \cup ( A\cap B^{c})

where A and A ∩ Bc  are disjoint.

P(A)=P(B\cup ( A\cap B^{c}))\\\\P(A)=P(B)+P( A\cap B^{c})

From axiom P(E)≥0

P( A\cap B^{c})\geq 0\\\\P(A)-P(B)=P( A\cap B^{c})\\P(A)=P(B)+P(A\cap B^{c})\geq P(B)

Therefore,

P(A)≥P(B)

8 0
3 years ago
Directions: Classify if the given mathematical symbols illustrate equation or
marissa [1.9K]

Answer:

Inequalities if it has \leq or \geq \\ symbols

Equations if it has =

Step-by-step explanation:

5 0
2 years ago
Central Park is a rectangular-shaped park in the middle of New York City. The area of the park is 3.41 square kilometers.
dmitriy555 [2]

Answer:

4,262.50 m

842 acres

Step-by-step explanation:

Given that :

Park is rectangular in shape

Width of park = 800 m

Area of park = 3.41 km²

Width of park = 800 / 1000 = 0.8 km

Area of rectangle, A

A = l * w

l = Length ; w = width

3.41 = l * 0.8

3.41 = 0.8l

l = 3.41 / 0.8

l = 4.2625 km

Hence, length in meters :

4.2625 * 1000

= 4,262.50 m

B.)

1 km² is approximately 247 acres

3.41km² = 3.41 * 247

3.41 km² = 842.27

= 842 acres

6 0
2 years ago
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