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7 months agoNew answer posted
7 months agoContributor-Level 10
Re (z²) = x²-y². 2 (Im (z)² = 2y². 2Re (z) = 2x.
x²-y²+2y²+2x=0 ⇒ x²+y²+2x=0.
(x+1)²+y²=1. Center (-1,0).
Parabola: x²-6x+9 = y-13+9 ⇒ (x-3)²=y-4. Vertex (3,4).
Line through (-1,0) and (3,4): slope = (4-0)/ (3- (-1) = 1.
y-0 = 1 (x+1) ⇒ y=x+1.
y-intercept is 1.
New answer posted
7 months agoContributor-Level 10
dy/dx = 2y/ (xlnx).
dy/y = 2dx/ (xlnx).
ln|y| = 2ln|lnx| + C.
ln|y| = ln (lnx)²) + C.
y = A (lnx)².
(ln2)² = A (ln2)². ⇒ A=1.
y = f (x) = (lnx)².
f (e) = (lne)² = 1² = 1.
New answer posted
7 months agoContributor-Level 10
P (at least one head) = 1 - P (no heads) = 1 - (1/2)? ≥ 0.9.
0.1 ≥ (1/2)?
10 ≤ 2?
n=3, 2³=8. n=4, 2? =16.
Minimum value of n is 4.
New answer posted
7 months agoContributor-Level 10
Sum S = Σ (2r+1)? C? = 2Σr? C? + Σ? C? = 2n2? ¹ + 2? = n2? + 2? = (n+1)2?
(n+1)2? = 101 * 2¹? ⇒ n=100.
2 [ (n-1)/2] = 2 [ (99)/2] = 2 [49.5] = 2*49 = 98.
New answer posted
7 months agoContributor-Level 10
(a+3b). (7a-5b) = 7|a|² - 5ab + 21ab - 15|b|² = 7|a|²+16ab-15|b|²=0.
(a-4b). (7a-2b) = 7|a|² - 2ab - 28ab + 8|b|² = 7|a|²-30ab+8|b|²=0.
Subtracting: 46ab - 23|b|² = 0 ⇒ 2ab = |b|².
Substituting: 7|a|² + 8|b|² - 15|b|² = 0 ⇒ 7|a|² = 7|b|² ⇒ |a|=|b|.
cosθ = ab/ (|a|b|) = ab/|b|² = (1/2)|b|²/|b|² = 1/2.
θ = 60°.
New answer posted
7 months agoContributor-Level 10
2log? (2? -5) = log?2 + log? (2? -7/2).
(2? -5)² = 2 (2? -7/2) = 2*2? -7.
Let t=2? (t-5)² = 2t-7.
t²-10t+25=2t-7 ⇒ t²-12t+32=0 ⇒ (t-4) (t-8)=0.
t=4 or t=8.
2? =4 ⇒ x=2. log? (4-5) undefined.
2? =8 ⇒ x=3. log? (8-5)=log?3=1. log? (8-3.5)=log?4.5.
2 (1) = log?2+log?4.5 = log?9=2. Correct.
x=3.
New answer posted
7 months agoContributor-Level 10
log? (18x-x²-77)>0 ⇒ 18x-x²-77>1 ⇒ x²-18x+78<0. Roots are 93.
log? (.)>0 ⇒ log? (.)>1 ⇒ 18x-x²-77>3 ⇒ x²-18x+80<0 (x-8) (x-10)<0.
8
I = ∫? ¹? sin³x/ (sin³x+sin³ (18-x)dx. Using King's property.
I = ∫? ¹? sin³ (18-x)/ (sin³ (18-x)+sin³x)dx.
2I = ∫? ¹? dx = 2. I=1.
New answer posted
7 months agoContributor-Level 10
f (1)=1.
f (4)=f (2)²=1 or 4.
f (6)=f (2)f (3).
Possible functions determined by values at primes: f (2), f (3), f (5), f (7).
f (2) can be 1 or 2. f (3) can be 1 or 3. f (5)=1,5. f (7)=1,7.
If f (m)=m, f (mn)=mn. One function. f (x)=1 is another.
What if f (2)=1, f (3)=3? f (6)=3.
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