Answer:
(x+3)(4x−2) = 4x2+10x−6
(x+2)(x2−3x +4) = x3−x2−2x+8
Step-by-step explanation:
Step-by-step explanation:
so basically they are asking you to subtract, hope this helps, use a calculator if needed
Define unit vectors along the x-axis and the y-axis as
![\hat{i} , \, \hat{j}](https://tex.z-dn.net/?f=%5Chat%7Bi%7D%20%2C%20%5C%2C%20%5Chat%7Bj%7D)
respectively.
Then the vector from P to Q is
![\vec{PQ} = (-13+5)\hat{i} + (10-5)\hat{j} = -8\hat{i} + 5\hat{j}](https://tex.z-dn.net/?f=%5Cvec%7BPQ%7D%20%3D%20%28-13%2B5%29%5Chat%7Bi%7D%20%2B%20%2810-5%29%5Chat%7Bj%7D%20%3D%20-8%5Chat%7Bi%7D%20%2B%205%5Chat%7Bj%7D)
In component form, the vector PQ is (-8,5).
The magnitude of vector PQ is
√[(-8)² + 5²] = √(89) = 9.434
Answer:
The vector PQ is (-8, 5) and its magnitude is √89 (or 9.434).
Answer: A.
Step-by-step explanation:
Answer:
The circle's centre is at the position (3, 5), and it has a radius of 2
Step-by-step explanation:
First let's put it in a useful format by completing the squares:
x² + y² - 6x - 10y + 30 = 0
x² - 6x + y² - 10y = -30
x² - 6x + 9 + y² - 10y + 25 = -30 + 9 + 25
(x - 3)² + (y - 5)² = 4
This tells us that the centre position is (3, 5) and the radius is √4, or 2