Answer:
<h2>A. (0,1)</h2>
Step-by-step explanation:
The question lacks the e=required option. Find the complete question below with options.
Which of the following points does not belong to the quadratic function
f(x) = 1-x²?
a.(0,1) b.(1,0) c.(-1,0)
Let f(x) = 0
The equation becomes 1-x² = 0
Solving 1-x² = 0 for x;
subtract 1 from both sides;
1-x²-1 = 0-1
-x² = -1
multiply both sides by minus sign
-(-x²) = -(-1)
x² = 1
take square root of both sides;
√x² = ±√1
x = ±1
x = 1 and x = -1
when x = 1
f(x) = y = 1-1²
y = 1-1
y = 0
when x = -1
f(x) = y = 1-(-1)²
y = 1-1
y = 0
Hence the coordinate of the function f(x) = 1-x² are (±1, 0) i.e (1, 0) and (-1, 0). The point that does not belong to the quadratic function is (0, 1)
Answer:
Vota Trump
Step-by-step explanation:
Answer:
look at the picture i have sent
(24xy^3-16x^2y^2+32x^2y)/8xy
<span><span>(<span><span><span><span><span><span><span>24x</span><span>y^3</span></span>−<span><span>16<span>x^2</span></span><span>y^2</span></span></span>+<span><span>32<span>x^2</span></span>y</span></span>8</span></span>x</span>)</span><span>(y)</span></span><span> =<span><span><span>−<span><span>2<span>x^3</span></span><span>y^3</span></span></span>+<span><span>3<span>x^2</span></span><span>y^4</span></span></span>+<span><span>4<span>x^3</span></span><span>y^<span>2</span></span></span></span></span>
