The answer would be 10 to the third because if you divide that it would equal 1,000
Answer:in what ?
Step-by-step explanation:
Answer:
Step-by-step explanation:
Okay, so I think I know what the equations are, but I might have misinterpreted them because of the syntax- I think when you ask a question you can use the symbols tool to input it in a more clear way, otherwise you can use parentheses and such.
Problem 1:
(x²)/4 +y²= 1
y= x+1
*substitute for y*
Now we have a one-variable equation we can solve-
x²/4 + (x+1)² = 1
x²/4 + (x+1)(x+1)= 1
x²/4 + x²+2x+1= 1
*subtract 1 from both sides to set equal to 0*
x²/4 +x^2+2x=0
x²/4 can also be 1/4 * x²
1/4 * x² +1*x² +2x = 0
*combine like terms*
5/4 * x^2+2x+ 0 =0
now, you can use the quadratic equation to solve for x
a= 5/4
b= 2
c=0
the syntax on this will be rough, but I'll do my best...
x= (-b ± √(b²-4ac))/(2a)
x= (-2 ±√(2²-4*(5/4)*(0))/(2*(5/4))
x= (-2 ±√(4-0))/(2.5)
x= (-2±2)/2.5
x will have 2 answers because of ±
x= 0 or x= 1.6
now plug that back into one of the equations and solve.
y= 0+1 = 1
y= 1.6+1= 2.6
Hopefully this explanation was enough to help you solve problem 2.
Problem 2:
x² + y² -16y +39= 0
y²- x² -9= 0
Answer:
5
Step-by-step explanation:
because it is asking what minus 9 is equal to -4
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)
