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Andrews [41]
3 years ago
15

A rectangle is inscribed with its base on the x-axis and its upper corners on the parabola y = 7 − x 2 y=7-x2. what are the dime

nsions of such a rectangle with the greatest possible area?

Mathematics
1 answer:
slega [8]3 years ago
5 0
A rectangle is inscribed with its base on the x-axis and its upper corners on the parabola y = 7 − x 2 y=7-x2. what are the dimensions of such a rectangle with the greatest possible area?

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10

Step-by-step explanation:

50 plus 10 windows by 2 dollars will give you 70 dollars.

20 plus 10 Windows by 5 dollars give you 70 dollars

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Fill in the blank. Given O below, you can conclude that OD is congruent to ___________.
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The circle with center O has two chords AC and EF which are of same length 9.07.

OD and OB are the two perpendiculars drawn from the center O to the two chords AC and EF .It represents the distance of the chords from the centre.

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OD is congruent to OB.

Option A is the right answer.

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Let A, B, C and D be sets. Prove that A \ B and C \ D are disjoint if and only if A ∩ C ⊆ B ∪ D
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Step-by-step explanation:

We have to prove both implications of the affirmation.

1) Let's assume that A \ B and C \ D are disjoint, we have to prove that A ∩ C ⊆ B ∪ D.

We'll prove it by reducing to absurd.

Let's suppose that A ∩ C ⊄ B ∪ D. That means that there is an element x that belongs to A ∩ C but not to B ∪ D.

As x belongs to A ∩ C, x ∈ A and x ∈ C.

As x doesn't belong to B ∪ D, x ∉ B and x ∉ D.

With this, we can say that x ∈ A \ B and x ∈ C \ D.

Therefore, x ∈ (A \ B) ∩ (C \ D), absurd!

It's absurd because we were assuming that A \ B and C \ D were disjoint, therefore their intersection must be empty.

The absurd came from assuming that A ∩ C ⊄ B ∪ D.

That proves that A ∩ C ⊆ B ∪ D.

2) Let's assume that A ∩ C ⊆ B ∪ D, we have to prove that A \ B and C \ D are disjoint (i.e.  A \ B ∩ C \ D is empty)

We'll prove it again by reducing to absurd.

Let's suppose that  A \ B ∩ C \ D is not empty. That means there is an element x that belongs to  A \ B ∩ C \ D. Therefore, x ∈ A \ B and x ∈ C \ D.

As x ∈ A \ B, x belongs to A but x doesn't belong to B.  

As x ∈ C \ D, x belongs to C but x doesn't belong to D.

With this, we can say that x ∈ A ∩ C and x ∉ B ∪ D.

So, there is an element that belongs to A ∩ C but not to B∪D, absurd!

It's absurd because we were assuming that A ∩ C ⊆ B ∪ D, therefore every element of A ∩ C must belong to B ∪ D.

The absurd came from assuming that A \ B ∩ C \ D is not empty.

That proves that A \ B ∩ C \ D is empty, i.e. A \ B and C \ D are disjoint.

7 0
3 years ago
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