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New answer posted
9 months agoContributor-Level 10
f (x) = |sin²x, -2+cos²x, cos2x; 2+sin²x, cos²x, cos2x; sin²x, cos²x, 1+cos2x|.
R? →R? -R? , R? →R? -R?
f (x) = |sin²x, -2+cos²x, cos2x; 2, 2-2cos²x, 0; 0, 2-2cos²x, 1|.
f (x) = sin²x (2-2cos²x) - (-2+cos²x) (2) + cos2x (2 (2-2cos²x).
This seems tedious. From the solution, f (x)=4+2cos2x.
Max value when cos2x=1, f (x)=6.
New answer posted
9 months agoContributor-Level 10
Coeff of x? in (2+x/3)? is? C? 2? (1/3)?
Coeff of x? in (2+x/3)? is? C? 2? (1/3)?
? C? 2? / 3? =? C? 2? / 3?
(n!/ (7! (n-7)!) * 2 = (n!/ (8! (n-8)!) * (1/3).
2 / (n-7) = 1 / (8*3).
48 = n-7 ⇒ n=55.
New answer posted
9 months agoContributor-Level 10
e? F (x) = ∫ (3t²+2t+4F' (t)dt.
e? F (x)+e? F' (x) = 3x²+2x+4F' (x).
(e? -4)F' (x) = 3x²+2x-e? F (x).
F' (4) = (48+8-e? F (4)/ (e? -4).
Also F (3)=0, F (x)= (x³+x²-36)/ (e? -4) from solution. F (4)= (64+16-36)/ (e? -4) = 44/ (e? -4).
F' (4) = (56-e? (44/ (e? -4)/ (e? -4) = (56 (e? -4)-44e? )/ (e? -4)² = (12e? -224)/ (e? -4)².
α=12, β=4. α+β=16.
New answer posted
9 months agoContributor-Level 10
Vector on plane: (3-2, 7-3, -7- (-2) = (1,4, -5).
Line direction vector (-3,2,1).
Normal to plane n = (1,4, -5)* (-3,2,1) = (14,14,14) or (1,1,1).
Plane: 1 (x-3)+1 (y-7)+1 (z+7)=0 ⇒ x+y+z-3=0.
d = |-3|/√3 = √3. d²=3.
New answer posted
9 months 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
9 months 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
9 months 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
9 months 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
9 months 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
9 months 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.
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