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)
Answer:
n/a
Step-by-step explanation:
it just needs more work on the vertex.
the answer should be -4
hope this is correct for you and this helped
Answer:
<u>Residue</u>
Step-by-step explanation:
Let a and b be integers. We define a mod b to be the residue of dividing a by b. For example, if a evenly divides b, then a mod b=0, 20 mod 6= 2. The modulus operator is widely used in programming, and it is convenient when a and b are large numbers.
a mod b is always a nonnegative integer. In fact, 0≤ a mod b≤ |b-1| by the division algorithm. a and b can also be negative integers. Since 8=-(-5)+3 then 8 mod -5= 3.
As a final example, some known properties can be rewritten in terms of mod. a mod 2=0 if and only if a is even. a mod 2=1 if and only if a is odd.
Answer:
Step-by-step explanation:
<h3>Given</h3>
<h3>Find</h3>
- Solve for x
- Find x when p = -5
<h3>Solution</h3>
- 4(px + 1) = 64
- 4(px + 1)/4 = 64/4
- px + 1 = 16
- px = 15
- x = 15/p
<u>When p = -5, substitute p:</u>
