This is a quadratic equation. There's a shortcut (two shortcuts really) called the Shakespeare Quadratic Formula (2b or -2b) applicable when the middle term is even:
OK, our problem is
Typically we'd write 
but the answers here don't so we won't either.
Answer: C and D
<span>(5,-1)
The key to this problem is ignoring all the information that you don't need. You really don't care about the coordinates for T, U, or V. Just W is all that matters. So make a graph, label the X and Y axis, and put a point at (1,5). Now simply turn the entire paper clockwise 90 degrees. If you do this, you'll see that what was the X axis is now the Y axis. And what was the Y axis is now the X axis. Label the axis again, and read the location of the point you marked. If you do so, you'll see that you have to go 5 spaces along the X axis, hence the 5. And you'll go 1 space down on the Y axis, hence the -1.
So the coordinates of W(1,5) rotated 90 degrees clockwise around (0,0) changes to (5,-1).</span>
Assume (a,b) has a minimum element m.
m is in the interval so a < m < b.
a < m
Adding a to both sides,
2a < a + m
Adding m to both sides of the first inequality,
a + m < 2m
So
2a < a+m < 2m
a < (a+m)/2 < m < b
Since the average (a+m)/2 is in the range (a,b) and less than m, that contradicts our assumption that m is the minimum. So we conclude there is no minimum since given any purported minimum we can always compute something smaller in the range.
Answer:
2. 7
Step-by-step explanation:
Answer:
1
Use the quadratic formula
=
−
±
2
−
4
√
2
x=\frac{-{\color{#e8710a}{b}} \pm \sqrt{{\color{#e8710a}{b}}^{2}-4{\color{#c92786}{a}}{\color{#129eaf}{c}}}}{2{\color{#c92786}{a}}}
x=2a−b±b2−4ac
Once in standard form, identify a, b, and c from the original equation and plug them into the quadratic formula.
2
+
5
−
2
=
0
x^{2}+5x-2=0
x2+5x−2=0
=
1
a={\color{#c92786}{1}}
a=1
=
5
b={\color{#e8710a}{5}}
b=5
=
−
2
c={\color{#129eaf}{-2}}
c=−2
=
−
5
±
5
2
−
4
⋅
1
(
−
2
)
√
2
⋅
1
Step-by-step explanation:
this should help
