Answer:
Step-by-step explanation:
![(x+1-i)(x+1+i)\\\\=(x+1)^2 -i^2~~~~~~~~~~~~~;[a^2 -b^2 = (a+b)(a-b)]\\\\=x^2 +2x +1 -(-1)\\\\=x^2 +2x +1+1\\\\=x^2 +2x +2](https://tex.z-dn.net/?f=%28x%2B1-i%29%28x%2B1%2Bi%29%5C%5C%5C%5C%3D%28x%2B1%29%5E2%20-i%5E2~~~~~~~~~~~~~%3B%5Ba%5E2%20-b%5E2%20%3D%20%28a%2Bb%29%28a-b%29%5D%5C%5C%5C%5C%3Dx%5E2%20%2B2x%20%2B1%20-%28-1%29%5C%5C%5C%5C%3Dx%5E2%20%2B2x%20%2B1%2B1%5C%5C%5C%5C%3Dx%5E2%20%2B2x%20%2B2)
Always find the difference in prices first.
<span>3.45 - 3.20 = $.25 difference </span>
<span>Always divide the difference by the original price. Change this number into a per cent. </span>
<span>.25/3.45 = .07246 = 7.2% decrease </span>
Answer:
p = 5
Step-by-step explanation:
cross-multiply: 9 = p + 4
p = 5
Answer:
The value of A is 5
Step-by-step explanation:
- The number is divisible by 3 if the sum of its digits is a number
divisible by 3
- Ex: 126 is divisible by 3 because the sum of its digits = 1 + 2 + 3 = 6
and 6 is divisible by 3
- The number is divisible by 5 if its ones digit is zero or 5
- Ex: 675 is divisible by 5 because its ones digit is 5
890 is divisible by 5 because its ones digit is 0
- We are looking for the value of A in the 4-digit number 3A5A which
makes the number divisible by both 3 and 5
∵ A is in the ones position
∴ A must be zero or 5
- Let us try A = 0
∵ A = 0
∴ The number is 3050
∵ The sum of the digits of the number = 3 + 0 + 5 + 0 = 8
∵ 8 is not divisible by 3
∴ 3050 is not divisible by both 3 and 5
∴ A can not be zero
- Let us try A = 5
∵ A = 5
∴ The number is 3555
∵ The sum of the digits of the number = 3 + 5 + 5 + 5 = 18
∵ 18 is divisible by 3
∴ 3555 is divisible by both 3 and 5
∴ A must be equal 5
* <em>The value of A is 5</em>
Answer: Choice D
=====================================================
Explanation:
The left portion is the interval (-∞, -2)
This is a shorthand way of saying 
The curved parenthesis says "do not include this endpoint as part of the solution set". Note the open hole at x = -2 in the diagram.
In contrast, the value x = 4 is included (due to the filled in circle), so we use a square bracket for this endpoint. Therefore, the right-hand portion is represented by [4, ∞) which translates to 
Negative and positive infinity will always use a parenthesis, and never a square bracket. This is because we can only approach infinity but never reach it, so we cannot include it as an endpoint.
All of this builds up to the full interval notation to be
The only square bracket is near the 4; everything else is a curved parenthesis. This is why choice D is the final answer.
