Answer:
pick the value that when you solve it, you get the correct answer
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
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Step-by-step explanation: there u go
Why not draw the graph of y=|x|? It has a v-shape, and the vertex is at (0,0).
The graph of f(x) = |x| - 3 looks exactly the same, EXCEPT that the whole graph of |x| is shifted 3 units downward. The smallest value that y can have is therefore -3.
Can you now figure out the range of f(x)?
Answer:
see explanation
Step-by-step explanation:
Since p(x) = x² + 2 , if x ≤ 4
For x = 4 then this is included in the inequality x ≤ 4
whereas x > 4 does not include x = 4 but values greater than 4
Thus to evaluate x = 4 use p(x) = x² + 2