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:
All real numbers are solutions
Step-by-step explanation:
Let's solve your equation step-by-step.
−2(2x+3)=−4(x+1)−2
Step 1: Simplify both sides of the equation.
−2(2x+3)=−4(x+1)−2
(−2)(2x)+(−2)(3)=(−4)(x)+(−4)(1)+−2(Distribute)
−4x+−6=−4x+−4+−2
−4x−6=(−4x)+(−4+−2)(Combine Like Terms)
−4x−6=−4x+−6
−4x−6=−4x−6
Step 2: Add 4x to both sides.
−4x−6+4x=−4x−6+4x
−6=−6
Step 3: Add 6 to both sides.
−6+6=−6+6
0=0
Answer:
All real numbers are solutions.
I'm guessing this is a parallelogram or something, so VR=TS and RS=VT
therefore, y=x+11 and x+2=y-3x
y=x+11 , 2=y-4x
y=x+11 , y=4x+2
then use the process of elimination:
y=4x+2 then plug in: y=3+11
- y=x+11 y=14
0=3x-9 I think this is how you'd do it, tbh it's just an educated
3x=9 , x=3 guess. Btw, sorry if it looks weird
Answer:
u=-1
Step-by-step explanation:
-3|2-4u|+5<-13
-6+12u+5<-13
12u<-13-5+6
12u<-12
12u/12<-12/12
u=-1
Answer:
I would say C
Step-by-step explanation:
Since two negatives make a positive
