I am really having a hard time with these 2 problems Please help me!!Which expression is evaluated first?All we need do is find constant k and rewrite this substituting what we get for constant k To get k, we substitute the data given, x=3, y=12, z=2 Now we rewrite substituting for constant Simplify by multiplying top and bottom on the right side by 2 Edwin
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2^x=3^y=12^z
2^x=3^y=12^z-2 x = 3 y = 1 2 z \large 2^{x} = 3^{y} = 12^{z} 2 x = 3 y = 1 2 z If the equation above is fulfilled for nonzero values of x , y , z , x,y,z, x , y , z , find the value of zIf 2^x = 3^y = 12^z , then find the value of z (x 2y)xy
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Hi I'm Piyush I'm obsessed by Maths It's my drug And here's your solution Solution 2 x = 3 y = (12) z = k 2 x = k, 3 y = k, (12) z = k 2 = k 1/x, 3 = k 1/y, 12 = k 1/z 12 = 2 2 x 3 k1/z = (k 1/x) 2 x k 1/y k1/z = k 1/x x 2 x k 1/ Since, (a m) n = a mn k1/z = k 2/x x k 1/y k1/z = k 2/x1/y Since, am x an = amn 1/z = 2/x 1/y, Since, bases are same ie base = k)Solve any way you can a) 2x y z = 8 x 2y z = 3 x 2y z = 7 Add the 1st and 3rd equations and the z's will cancel out 2x y z = 8 x 2y z = 7 3x 3y = 15 And what's more, that can be divided through by 3 x y = 5 Next add the 2nd and 3rd equations andHence, the maximum value of Z is 18 at the point (4, 3) 6 Minimise Z = x 2y subject to Solution The feasible region determined by the constraints, 2x y ≥ 3, x 2y ≥ 6, x ≥ 0, and y ≥ 0, is given below A (6, 0) and B (0, 3) are the corner points of the feasible region The values of Z at the corner points are given below
If 2^x = 3^y =12^z show that 1/z= 1/y2/x Watch later Share Copy link Info Shopping Tap to unmute If playback doesn't begin shortly, try restarting your device Up nextI collected a solution Need to prove x^2y^2z^25xyz \ge 8 Put xyz=p;xyyzzx=q;xyz=r We have qr=4 Need to prove inequality is equivalent toA6 b8 c10 d12 C The world population is over 7 billion Which declaration uses the fewest bits while guaranteeing that worldPopulation can be assigned the value 7 billion without error?
2 x = 3 y = 12 z Let 2 x = 3 y = 12 z = c where c is some constant Now using the relations, 2 x = c x = log 2 c And, 3 y = c y = log 3 c Therefore, 1 y 2 x = 1 log 3 c 2 l o g 2 c Using 1 log a b = log b a, 1 y 2 x = log c 3 2 log c 2 1 y 2 x = log c 3 log c 2 2 1 y 2 x = log c 3 log c 4 1 y 2 x = log c (3 ⋅ 4) 1 y 2 x = log c 12 And since, log c 12 = 1 log 12 c = 1 z We have, 1X – 2y z = 3 − 3x 6y – 3z = − 9 4x – 8y 4z = 12 For the first step, you would choose two equations and combine them to eliminate a variable You can eliminate x by multiplying the first equation by 3 and adding to the second equationView Homework Help EE07 Tut1 Solution from EE 07 at Nanyang Technological University EE07 Tutorial 1 1i) 3 x 2 y 3 z 2 x 3 y z 3 x 2 y 3z 1 E1 x 2 y 3z 1 y 4 z 2 8 y 12 z 5 E2 E2 E1
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FX,Y (x,y) = ˆ 2 x ≥ 0,y ≥ 0,x y ≤ 1, 0 otherwise What is the variance of W = X Y?•2(x3)=2xλ (12) •2(y1)=2yλ (13) •2(z1)=2zλ (14) •x 2y z2=4 (15) •The simplest way to solve these equations is to solve for x, y, and z in terms of λ from (12), (13), and (14), and then substitute these values into (15) Class 12 Maths (with MCQs) NCERT Solutions Science;
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Z varies jointly with y and x, and z = 192 when x = 2 and y = 6, solve for z when b = 2,c = 3 The statement denotes a relationship of z = kyx where k is a constant Plugging in our initial statement values of z = 192 when x = 2 and y = 6, we get 192 = (2)(6)k 192 = 12k Divide each side byE 2 x y 12 z = 1 x 2 y 9 z = − 1 is a line in R 3 , as we saw in this example These equations are called the implicit equations for the line the line is defined implicitly as the simultaneous solutions to those two equationsWe know that, ( a m) n = a m n a m × a n = a m n a m ÷ a n = a m − n a 0 = 1 Therefore, Let 2 x = 3 y = 12 z = k This implies, 2 = k 1 x, 3 = k 1 y, 12 = k 1 z
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Problem 614 Solution We can solve this problem using Theorem 62 which says that VarW = VarX VarY 2CovX,Y (1) The first two moments of X are EX = Z 1 0 Z 1−x 0 2xdydx = Z 1 0 2x(1 −x)dx = 1/3 (2) E X2 = Z 1 0 Z 1−x 0 2x2 dydx = Z 12 x = 3 y = 1 2 z \large 2^ {x} = 3^ {y} = 12^ {z} 2x=3y=12z If the equation above is fulfilled for nonzero values of x, y, z, x,y,z, x,y,z, find the value of zAnswer with step by step detailed solutions to question from HashLearn's Math (ICSE 9), Number Systems "If 2^x=3^y=12^z then the value of x is in terms of y and z is" plus 1918 more questions from Mathematics Questions of this type are frequently asked in competitive entrance exams like ICSE Class 9 and
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∂ f ∂ y x, z = x 4 12 x 3 y 4 x y 2 z 6 y 2 correct incorrect ∂ f ∂ y x , z = 17 x 4 4 x 2 y z 12 x y correct incorrect * not completedClass 9Exponents of real numbersRD Sharma Pg 225 Q 5 If 2^x=3^y=12^z, show that 1/z=1/y2/xLaws of Exponents1 a ^n ⋅ a ^m = a^( nm)2 a ^n / a^ m = a^ (Class 8 Maths Chapter 14 Extra Questions Factorise (а) 14m 5 n 4 p 2 – 42m 7 n 3 p 7 – 70m 6 n 4 p 3 (b) 2a 2 (b 2 – c 2) b 2 (2c 2 – 2a 2) 2c 2 (a 2 – b 2) The area of a rectangle is 6a 2 36a and 36a width Find the length of the rectangle What are the common factors of
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2^x = 3^y = 12^z = k(let) then, 2^x=k 2 = k^1/x 3^y=k 3 = k^1/y and 12^z=k 12 = k^1/z or, 3×4 = k^1/z or, k^1/y × 2 x 2 = k^1/z or, k^1/y × k^1/x × k^1/x = k^1/z or, k^1/y1/x1/x = k^1/z or, k^1/y2/x = k^1/z or, 1/y2/x = 1/z or, x2y/xy = 1/z or, x2y = 1/z ×xy or, x2y = xy/z or, z(x2y) = xy or, zx 2yz = xy or, 2yz = xyzx or, 2yz = x(yz) or, x(yz) = 2yz The first, λ = 0 λ = 0 is not possible since if this was the case equation (1) (1) would reduce to y z = 0 ⇒ y = 0 or z = 0 y z = 0 ⇒ y = 0 or z = 0 Since we are talking about the dimensions of a box neither of these are possible so we can discount λ = 0 λ = 0 This leaves the second possibility x z = y z x z = y z2 4 x 1 x 2 x 3 3 5 2 Example Let T(x;y;z) = x 2y;2x y Since T
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Corresponding elements are equal 2x 3 = 9 2y = 12 2z – 3 = 15 2t 6 = 18 Solving equation (1) 2x 3 = 9 2x = 9 – 3 x = 6/2 x = 3 Solving equation (2) 2y = 12 y = 12/2 y = 6 Hence x = 3 , y = 6 , z = 9 & t = 6 Show MoreY y varies jointly as the square of x x and the square of z z and when x = 3 x = 3 and z = 4, z = 4, then y = 72 y = 72 19 y y varies jointly as x x and the square root of z z and when x = 2 x = 2 and z = 25 , z = 25 , then y = 100 y = 100Evaluate the following java expression,if x=3,y=5 and z=10 zyyzx Posted on by Views Score Share EngineeringCS EngineeringIS mca JIT Davangere SEMVI Java
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Write the following expressions in terms of logs of x, y and z (1) logx2y (2) log x3y2 z (3) log p x 3 p y2 z4 (4) logxyz (5) log x yz (6) log x y 2 (7) log(xy) 1 3 (8) logx p z (9) log 3 p x 3 p yz (10) log 4 r x3y 2 z4 (11) logx rp x z (12) log r xy z8 4 Write the following equalities in exponential form (1) log 3 81 = 4 (2) log 7 7 = 1 The answer is =6 Firstly, xyyzzx=12x^2 =>, xyx^2yzzx=12 =>, x(xy)z(xy)=12 =>, (xy)(xz)=12 Therefore, {(xy=Z x x y − − 6 3 if x = 2, y = 8, and z = –2 ( ) ( ) 1/7 14 2 122 6 8 3 2 8 = − − − − 6 −2 − 5 B Simplify ( ) 2 4 14 30 − − ( ) 2 8 16 2 4 14 30 = − − − − 6 C Use the distributive property to simplify −3(x−10) x ( ) 2 30 3 30 3 10 − − − − x x x x x 7 D Simplify 8y – 2 – 3(y – 4
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