[ Answer ]
[ Explanation ]
- System Of Equations
-----------------------------------
- Isolate x for x + x + 10 = 10: x = 0
For y = 10
Substitute x = 0
Y = 0 + 10
0 + 10 = 10
y = 10
X = 0
Y = 10
![\boxed{\bold{[] \ Eclipsed \ []}}](https://tex.z-dn.net/?f=%5Cboxed%7B%5Cbold%7B%5B%5D%20%5C%20Eclipsed%20%5C%20%5B%5D%7D%7D)
Let p(x) be a polynomial, and suppose that a is any real
number. Prove that
lim x→a p(x) = p(a) .
Solution. Notice that
2(−1)4 − 3(−1)3 − 4(−1)2 − (−1) − 1 = 1 .
So x − (−1) must divide 2x^4 − 3x^3 − 4x^2 − x − 2. Do polynomial
long division to get 2x^4 − 3x^3 − 4x^2 – x – 2 / (x − (−1)) = 2x^3 − 5x^2 + x –
2.
Let ε > 0. Set δ = min{ ε/40 , 1}. Let x be a real number
such that 0 < |x−(−1)| < δ. Then |x + 1| < ε/40 . Also, |x + 1| <
1, so −2 < x < 0. In particular |x| < 2. So
|2x^3 − 5x^2 + x − 2| ≤ |2x^3 | + | − 5x^2 | + |x| + | − 2|
= 2|x|^3 + 5|x|^2 + |x| + 2
< 2(2)^3 + 5(2)^2 + (2) + 2
= 40
Thus, |2x^4 − 3x^3 − 4x^2 − x − 2| = |x + 1| · |2x^3 − 5x^2
+ x − 2| < ε/40 · 40 = ε.
Answer:
<QOS =150°(opp. <'s of a cyclic quard)
<QRS=75°(< at centre =2×< at circumference)
I don't know about the third one bro
Maybe there's an erroron that quation
Answer: Choice D) The h(x) values -4, -2, 2, 4, and 6
Explanation:
The range is the set of all possible y outputs, or h(x) outputs.
We simply list the y coordinate of each blue point shown on the graph.
We cannot say something like -4 ≤ h(x) ≤ 6 because there are gaps where the function is not defined. For instance, h(x) = 0 is not possible, so 0 is not in the range. Put another way: there is no point with y coordinate 0 on the graph, so that's why 0 is not part of the range.
