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X=1/2 y=2/3 z=-1/5-Z=1/5 Get the answers you need, now! y 5 2 1 2 3 y 2 3 5 2 x 2 3 y 2 3 5 2 x 2 3 y 1 1 5 5 7 5 or 14 d 5 34 1 4 2 from CALCULUS MCV4U at St Augustine Catholic High School, Tucson




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Y = 2 3t ;z = 3 2t` and the direction vector of the line is `bar v_1 = lt2,3,2gt`See answers gopi05 gopi05 gopi05 Download PDF's Class 12 Class 11 Class 10 Class 9 Class 8 Class 7 Class 6 NCERT Easy Reading Alleen Test Solutions Blog About Us Career
`(z3)/(2) = t =gt z = 3 2t` Hence, the parametric equations of the first line are `x = 1 2t ; Show that the lines `(x1)/(2)=(y1)/(3)=z " and " (x1)/(5)=(y2)/(1)z=2` class12;Solve your math problems using our free math solver with stepbystep solutions Our math solver supports basic math, prealgebra, algebra, trigonometry, calculus and more
Solve your math problems using our free math solver with stepbystep solutions Our math solver supports basic math, prealgebra, algebra, trigonometry, calculus and moreSolve your math problems using our free math solver with stepbystep solutions Our math solver supports basic math, prealgebra, algebra, trigonometry, calculus and moreLines (x 1)/2 = (y 2)/3 = (z 3)/4 and (x 4)/5 = (y 1)/2 = z are intersecting in ( ) ( 1, 1, 1) (D) ( 1, 1, 1)




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Let L 1 (x – 1)/2 = (y – 2)/3 = (z – 3)/4 and L 2 (x – 2)/3 = (y – 4)/4 = (z – 5)/5 and A = (2, 3, 5) The equation of the plane passing through A and containing the line L 1 is (a) 2x – z 1 = 0 (b) 2x – y – z 2 = 0 (c) 2x – y 3z 2 = 0 (d) 3x – y 5z 10 = 0Lines `(x1)/2 = (y2)/3 = (z3)/4` and `(x2)/3 = (y3)/4 = (z4)/5` lie on the planeThe distance of the point 1, –2, 3 from the plane x – y z = 5 measured parallel to the line 1/2x = 1/3y = 1/6z is




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(x 1)/3 = (y 1)/2 = (z 1)/5 and (x 2)/4 = (y 1)/3 = lines are ____ (A) parallel (B) coincident Intersecting (D) skewY = 3 x x2 3x 2 = 0 L'equazione di secondo grado presente nel sistema ammette come soluzioni x = 1 e x = 2, quindi il sistema risulta equivalente a 8 < y = 3 x x = 1 _ 8 < x = 2 le cui soluzioni sono 8 < y= 2 x = 1 _ 8 < = 1 x = 2 Esercizi Risolvere i seguenti sistemi 1 8 < x y = 2 3xy = 81 2 8 < 4x1 8y = 1 25x = 5 1252y 3 8The coordinates of any point on the line (2) is given as x=5s4 , y=2s1 , z=s(2)' now if the lines intersect , the intersection point will lie on both the equation 2t1=5s4 now solving (3) and (4) 3*(3)2*(4) from equation ;(3) 2t5=3 2t=2 t=1 now the zcoordinates of equation (1) is 4*(1)3=43=1 zcoordinates of equation



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The correct option is C 3 5 Y ∩ Z = ϕ ⇒ Y and Z are disjoint subsets of X Let a ∈ X Case 1 a ∈ Y and a ∉ Z Case 2 a ∈ Z and a ∉ Y Case 3 a ∉ Y and a ∉ Z So, each element has 3 options There are 5 possibilities for a ∴ Possible number of ordered pairs = 3 5The correct option is B 3 5 Y ∩ Z = ϕ ⇒ Y and Z are disjoint subsets of X Let a ∈ X Case 1 a ∈ Y and a ∉ Z Case 2 a ∈ Z and a ∉ Y Case 3 a ∉ Y and a ∉ Z So, each element has 3 options There are 5 possibilities for a ∴ Possible number of ordered pairs = 3 5X^2 y^2 z^2 = 1 Natural Language;




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Expert Answer let s2 be the unit sphere x2 y2 z =1 in p3 Defin in 52 The six charts corres ponding to the six hemispheres the fornt, right, left, upper and lower hemispheres Prove that charts are C ∞ on the inter sections of hemispheres (1) Prove that ϕ10ϕ4−1 is C ∞ϕ4(0,4) (2) Prove that ϕ1 ∘ϕ5−1 is e∞ on ϕ5 (U 15) (3Solution The correct option is A 1 Line parallel to the line x 2= y 2= z −6 and passing through (1,2, 3) is x−1 2 = y2 3 = z−3 −6 = r A = (2r 1, 3r – 2, – 6r 3) is a point on this line A lies in the plane x – y z = 5 ∴ 2r1−3r2−6r3 =5 ⇒r = 1 7 ∴ A=(9 7, −11 7, 15 7) Distance of A from (1, –2, 3) is Re funzione X = 1 1/2X , 1255 maffio ha scritto chi mi descrive i passaggi per ottenere come risultato x=2 di questa semplice funzione X = 1 1/2X ?



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1/x 1/y 1/z = 235 Get the answers you need, now!The correct option is A x3 3 = y−5 1 = z−2 −5 Here, plane , line and its image are parallel to each other So, find any point on the normal to the plane from which the image line will be passed and then find equation of image line Here, plane and line are parallel to each other Equation of normal to the plane through the point (1, 310 0 10 x40 0 40 y 0 25 50 75 100 z f(x,y)=x2 a2 y2 b2, PARABOLOIDE (ellittico) con a =2eb =3 Se a =1,b= 1 esso si ottiene dalla rotazione della curva z = y2 del piano zy attorno all'asse z10 5 0 5 10 x105 0 5 y 1010 0 10 z 10 f(x,y)=x2 a2 − y2 b2, PARABOLOIDE IPERBOLICO, sella dicavallo cona =2,b=3 Le intersezioni del grafico con piani y = c danno




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Show that the lines \(\cfrac{\text x1}3=\cfrac{y1}2=\cfrac{z1}5\) and \(\cfrac{\text x2}2=\cfrac{y1}3=\ , (x 2)/2=(y 1)/3 = (z 1)/2रेखा `(x4)/(5)=(y1)/(2)=(z)/(1)` तथा `(x1)/(2)=(y2)/(3)=(z3)/(4)` के बीच का कोण है2 days ago Fast and easy to use Worksheet 1 Read Sec 3 See More Examples » Free Calculus Worksheets to Download CONCEPT ONE Graphing exponential functions 4 3 11 12 = 11 9 =1 2 9 2 ANSWERS a) 2 (3) 4 y f x b) 1 3 2 y f x c) 1 2 (3(1)) 3 y f x d) (2 4) y f x PreCalculus is a rigorous course that requires a commitment to studying 38




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