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New answer posted
a year agoContributor-Level 10
(a+b+c)² = a²+b²+c²+2 (ab+bc+ca)
1² = a²+b²+c²+2 (2) ⇒ a²+b²+c² = -3.
a²b²+b²c²+c²a² = (ab+bc+ca)² - 2abc (a+b+c) = 2² - 2 (3) (1) = -2.
a? +b? +c? = (a²+b²+c²)² - 2 (a²b²+b²c²+c²a²) = (-3)² - 2 (-2) = 9+4=13.
New answer posted
a year agoContributor-Level 10
Δ = |1,1, -1; 1,2, α 2, -1,1| = 1 (2+α)-1 (1-2α)-1 (-1-4) = 2+α+2α-1+5 = 3α+6=0 ⇒ α=-2.
Δ? = |2,1, -1; 1,2, α β, -1,1| = 2 (2+α)-1 (1-αβ)-1 (-1-2β) = 4+2α-1+αβ+1+2β = 4+2α+αβ+2β=0.
4-4-2β+2β=0. This holds.
Δ? = |1,2, -1; 1,1, α 2, β,1| = 1 (1-αβ)-2 (1-2α)-1 (β-2) = 1-αβ-2+4α-β+2 = 1+4α-αβ-β=0.
1-8+2β-β=0 ⇒ -7+β=0 ⇒ β=7.
α+β = -2+7 = 5.
New answer posted
a year agoContributor-Level 10
sec y dy/dx = 2sinxcosy.
sec²y dy = 2sinx dx.
tan y = -2cosx + C.
y (0)=0 ⇒ 0=-2+C ⇒ C=2.
tan y = 2-2cosx.
y' = (-2sinx)/sec²y.
5y' (π/2) = 5 (2sin (π/2)/sec² (π/2)
sec²y dy/dx = 2sinx.
y' (π/2)? At x=π/2, tan y = 2. sec²y = 1+tan²y = 5.
5 (2sin (π/2) = 5 (2)=10.
New answer posted
a year agoContributor-Level 10
P (x) = a (x-2)² + b (x-2) + c.
lim (x→2) P (x)/sin (x-2) = lim (x→2) P (x)/ (x-2) = P' (2) = 7.
P' (x) = 2a (x-2) + b. P' (2) = b = 7.
P' (x) = 2a.
P (3) = a (1)² + b (1) + c = a+b+c = 9.
Continuity at x=2 means lim f (x) = f (2).
lim (x→2) (a (x-2)²+b (x-2)+c)/ (x-2) = P' (2) = b=7. This is given.
The problem states f (2)=7.
P (x) = (x-2) (ax+b) form used in solution. Let's use this.
lim (x→2) (x-2) (ax+b)/sin (x-2) = lim (x→2) ax+b = 2a+b = 7.
P (3) = (3-2) (3a+b) = 3a+b=9.
Solving: a=2, b=3.
P (x) = (x-2) (2x+3).
P (5) = (5-2) (2*5+3) = 3 * 13 = 39.
New answer posted
a year agoContributor-Level 10
Mean = (6+10+7+13+a+12+b+12)/8 = (60+a+b)/8 = 9 ⇒ a+b=12.
Variance = (Σx? ²/n) - (mean)² = 37/4.
Σx? ²/8 - 81 = 37/4.
Σx? ² = 8 (81+9.25) = 8 (90.25) = 722.
Σx? ² = 36+100+49+169+a²+144+b²+144 = 642+a²+b².
642+a²+b²=722 ⇒ a²+b²=80.
(a+b)²=144 ⇒ a²+b²+2ab=144 ⇒ 80+2ab=144 ⇒ 2ab=64.
(a-b)² = a²+b²-2ab = 80-64 = 16.
New answer posted
a year agoContributor-Level 10
x? + 2 (20)¹/? x³ + (20)¹/²x² + . No.
x² = - (20)¹/? x - (5)¹/².
α+β = - (20)¹/? , αβ=5¹/².
α²+β² = (α+β)²-2αβ = (20)¹/² - 2 (5)¹/² = 0.
α? +β? = (α²+β²)² - 2 (αβ)² = 0 - 2 (5) = -10.
α? +β? = (α? +β? )² - 2 (αβ)? = (-10)² - 2 (5)² = 100 - 50 = 50.
New answer posted
a year agoContributor-Level 10
Using L'Hopital Rule:
lim (x→2) (2xf (2) - 4f' (x)/1 = 2 (2)f (2) - 4f' (2) = 4 (4) - 4 (1) = 12.
New answer posted
a year agoContributor-Level 10
I = ∫? π/? π/? dx/ (1+e^ (xcosx) (sin? x+cos? x). Using ∫? f (x)dx = ∫? f (a+b-x)dx. a+b=0.
I = ∫? π/? π/? dx/ (1+e? ) (sin? x+cos? x) = ∫? π/? π/? e? dx/ (e? +1) (sin? x+cos? x).
2I = ∫? π/? π/? dx/ (sin? x+cos? x) = 2∫? π/? dx/ (sin? x+cos? x).
I = ∫? π/? sec? xdx/ (tan? x+1). Let t=tanx.
I = ∫? ¹ (t²+1)dt/ (t? +1) = ∫? ¹ (1+1/t²)dt/ (t²-√2t+1) (t²+√2t+1). No, this is hard.
I = ∫? ¹ (1+1/t²)dt/ (t-1/t)²+2). Let u=t-1/t. I = ∫ du/ (u²+2) = (1/√2)tan? ¹ (u/√2).
= π/ (2√2).
New answer posted
a year agoContributor-Level 10
C= (2,3), O= (0,0). r = OC = √13.
Slope of OC = 3/2. Slope of PQ = -2/3.
Let P= (x, y). Vector CP = (x-2, y-3). Vector OC = (2,3).
CP.OC=0 ⇒ 2 (x-2)+3 (y-3)=0 ⇒ 2x+3y=13.
Also (x-2)²+ (y-3)²=13.
From 2x=13-3y, x= (13-3y)/2.
(13-3y)/2-2)²+ (y-3)²=13 ⇒ (9-3y)/2)²+ (y-3)²=13
(9 (y-3)²/4) + (y-3)² = 13 ⇒ (13/4) (y-3)²=13 ⇒ (y-3)²=4 ⇒ y-3=±2.
y=5 or y=1.
If y=5, x=-1. P= (-1,5).
If y=1, x=5. Q= (5,1).
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