Answer:
Step-by-step explanation:
Y = -(1/2)(x-2)² +8
Re write it in standard form:
(y-8) = -1/2(x-2)² ↔ (y-k) = a(x-h)²
This parabola open downward (a = -1/2 <0), with a maximum shown in vertex
The vertex is (h , k) → Vertex(2 , 8)
focus(h, k +c )
a = 1/4c → -1/2 = 1/4c → c = -1/2, hence focus(2, 8-1/2) →focus(2,15/2)
Latus rectum: y-value = 15/2
Replace in the equation y with 15/2→→15/2 = -1/2(x-2)² + 8
Or -1/2(x-2)² +8 -15/2 = 0
Solving this quadratic equation gives x' = 3 and x" = 2, then
Latus rectum = 5
Answer:
titutex=cos\alp,\alp∈[0:;π]
\displaystyle Then\; |x+\sqrt{1-x^2}|=\sqrt{2}(2x^2-1)\Leftright |cos\alp +sin\alp |=\sqrt{2}(2cos^2\alp -1)Then∣x+
1−x
2
∣=
2
(2x
2
−1)\Leftright∣cos\alp+sin\alp∣=
2
(2cos
2
\alp−1)
\displaystyle |\N {\sqrt{2}}cos(\alp-\frac{\pi}{4})|=\N {\sqrt{2}}cos(2\alp )\Right \alp\in[0\: ;\: \frac{\pi}{4}]\cup [\frac{3\pi}{4}\: ;\: \pi]∣N
2
cos(\alp−
4
π
)∣=N
2
cos(2\alp)\Right\alp∈[0;
4
π
]∪[
4
3π
;π]
1) \displaystyle \alp \in [0\: ;\: \frac{\pi}{4}]\alp∈[0;
4
π
]
\displaystyle cos(\alp -\frac{\pi}{4})=cos(2\alp )\dotscos(\alp−
4
π
)=cos(2\alp)…
2. \displaystyle \alp\in [\frac{3\pi}{4}\: ;\: \pi]\alp∈[
4
3π
;π]
\displaystyle -cos(\alp -\frac{\pi}{4})=cos(2\alp )\dots−cos(\alp−
4
π
)=cos(2\alp)…
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You take all known values and subtract it from 360.
360-132-39-90=99.
The answer is 99.
30*30=900
hope this helps!
