Solve: a3z + 4. дхду? a3z a3z 4 dx²əy əx3

Q: Solve 2x-y=5 and 3x+2y=11 by using substitution method.

A: Given: 2x−y=5…

Q: . Solve: (6x + 1)y² + 3x² + 2y° %3D dr

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Q: Solve: x4 - 6x2 - 8x + 24 = 0.

A: Consider the given equation: Here the objective is to find the value of x.

Q: Solve f1x2 = 4x+1 - 2

A: f(x)=4x+1-2

Q: Solve by substitution: 10x=4 2x+y=-24

A: The given system of equations is: 10x = 4 2x + y = -24   The solution is below.

Q: Solve: 4x − 7y + 3z = 13x + y − 2z = 44x − 7y +3z =6

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Q: Solve using the substitution method. 9x-3y=12 and 3x-4=y

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Q: Solve using substitution y=2x+4 2x-y=-6

A: Given data: The expression for the first equation is y=2x+4. The expression for the second equation…

Q: Solve 4*- - gr-4 a. b. x- 2 d. 213

A: Solution: The objective is to solve the given equation

Q: Solve by substitution:  y = − 2x − 9y = − 5x − 21

A: We substitute y=-2x-9 in the other equation and solve for x

Q: Solve by substitution:  y = 2x − 3y = − 2x +9

A: To solve the given equations by substitution method.

Q: Solve by substitution:  2x + y = − 75x + 3y = − 21

A: Here, 2x + y = − 7 ….(1)5x + 3y = − 21 ….(2)

Q: solve for y  28= -6y-12+2y

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Q: V5 + 12x – 9x²dx

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Q: solve for the value of x: 10z-3=7z+9

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Q: Solve: -3z? – 6z – 2 = 5z2 – 10z – 10 Enter your answer in the form a + bi.

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Q: Solve for r  us = -rtv

A: Given,            us = -rtv

Q: solve for y. 4-y = 3y+2

A: To solve 4-y=3y+2

Q: Solve - 2 -ळ - 2z -3

A: Here, we need to solve 2z+1-2z-34=2-3i. Let us convert RHS to polar form. So, 2-3i=5cos-arctan32+i…

Q: Solve using the substitution method. 6x-3y=9 and 2x-3=y

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Q: Solve using substitution. y = -3x- 7 y = 9x + 5

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Q: Solve 71+3r9 +6xs- I+ I, = --2 211+ 8r2 + 16r3 - 3r4 +3x5 = -4 +20, + 4r3-3,+3x; -7

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Q: Solve by substitution:  y = − 2x + 8− 7x − 6y = − 8

A: Substitute y=-2x+8 in the other equation and solve for x

Q: +Solve by substitution y = 4x-3 and 3x+y=32 2x + 2y = 2 and x = y + 15

A: Given :y=4x-3        3x+y=32    2x+2y=2   x=y+15

Q: Solve by elimination  2x-6y=6 3x-y=25

A: We solve the given equation by elimination method and find x and y.

Q: Solve for y 6x + 2y = - 8

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Q: Solve using the substitution method.y=6-3x and 4x+y=6

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Q: Solve: y' + 4y - 17y' - 60y: = y(0) = - 4, y'(0) = - -5, y(t) = = - 72et y''(0) = - 65

A: First we will find the Auxiliary equation of given differential equation to get Complementry…

Q: Solve AXYZ where <Z = 90°, z= 10, & y= 6. Answers to 1 decimal place.

A: We will find the missing values of this triangle

Q: Solve by substitution:  y = 3x + 13y = − 2x − 22

A: Substitute y = 3x + 13 in the other equation and solve for x

Q: Solve for y 6x + 2y - 8

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Q: Solve by substitution:  y = − 6x + 3y = 6x +3

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Q: Solve for y: 2y + 3x - 5 = 4

A: Given 2y+3x-5=4

Q: Solve using the substitution method. y=2-5x 6x+y=6

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Q: solve  (-y)(y^2)

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Q: Q2.Solve IV. Por つイーの片+644€ ゴー2454-5メ+ 2+54-5xメナト

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Q: Solve:–2x+y=6 y = 3x + 4

A: the first equation is  -2x+y=6 substitute the value of y -2x+(3x+4)=6 -2x+3x+4=6

Q: Find all real solutions of the equation. 4z5 - 10z2 = 0

A: Given: 4z5-10z2=0  To Find all the real solution

Q: 3x+y%=14 x= 2y

A:

Q: 2 ( y' V(4-y') (4-у?)3

A: To evaluate : I=∫02y34-y23dy I=∫02y24-y23ydy Let 4-y2 = t ⇒-2ydy =dt ⇒ydy =dt-2 Now for, y=0, t=4…

Q: Solve ,  (D2-7D+12)y = 0

A: Given ,  (D2 - 7D + 12)y = 0 The auxiliary equation is  m2 - 7m + 12 = 0

Q: Solve using the substitution method.y=6-3x and 4x+y=12

A: Here we have to find the value of x and y.  Given Equation is y=6-3x ........................... (1)…

Q: Solve by substitution:  − 6x + y = 20− 3x − 3y = − 18

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Q: JAVAV Solve se , where y =(),-1, V= (0,4,3), and, w= (0,3, l1). 54-2x = 3 v+

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Q: solve. e4x-3e2x=18

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Q: Solve using the substitution method. 2x+4y=-3 and 2x-y=7

A: We have given2x+4y=-3               ⋯⋯⋯⋯(i)2x-y=7                     ⋯⋯⋯⋯(ii)

Q: Solve using the substitution method. y=2-5x and 6x+y=8

A: Explanation of the answer is as follows

Q: 40000y" - 81y = 0 y(0) = 4 , y'(0)= 9

A:

Equations and Inequations

Equations and inequalities describe the relationship between two mathematical expressions.

Linear Functions

A linear function can just be a constant, or it can be the constant multiplied with the variable like x or y. If the variables are of the form, x2, x1/2 or y2 it is not linear. The exponent over the variables should always be 1.