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Sloan [31]
3 years ago
12

Darren wins a coupon for $4 off the lunch special for each of five days he pays $75 for his 5 lunch specials write and solve an

equation to find the original price p four one lunch special.
P.s ur getting reported if you don't show all work and if you answer without an answer for the question
Mathematics
1 answer:
Valentin [98]3 years ago
8 0

The original price for one lunch special is $19.

<em><u>Explanation</u></em>

The original price for one lunch special is  'p'  dollar.

He wins a coupon for $4 off for each of five days. That means , <u>he needs to pay (p-4) dollar each day</u>.

So, the total amount needed to pay for 5 days = 5(p-4) dollar

Given that, <u>he pays $75 for his 5 lunch specials</u>. So the equation will be.....

5(p-4)= 75\\ \\ 5p-20=75\\ \\ 5p= 75+20\\ \\ 5p= 95\\ \\ p=\frac{95}{5}=19

So, the original price for one lunch special is $19.

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3.5,

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Which graph shows the solution set of the compound inequality or 1.5x-1 &gt; 6.5 or 7x+3 &lt; -25 ?
Dmitry [639]
1.
<span>1.5x-1 > 6.5 
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x>5, is represented by the region to the right of the vertical line x=5 

2. 

<span>7x+3 < -25
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2 years ago
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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.

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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)

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

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the 12 inch pizza is $1.72 per inch and the 14 inch pizza is $1.75 per inch. so the 14 inch pizza is the better choice.

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