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2 years ago

What is the the answer for 2(x - 4)=6x

Mathematics

2 answers:

anzhelika [568]2 years ago

7 0

Answer:

x = -2

Step-by-step explanation:

Step 1: Simplify both sides of the equation.

2(x−4)=6x

(2)(x)+(2)(−4)=6x (Distribute)

2x + −8 = 6x

2x − 8 = 6x

Step 2: Subtract 6x from both sides.

2x −8 − 6x = 6x − 6x

−4x − 8 = 0

Step 3: Add 8 to both sides.

−4x − 8 + 8 = 0 + 8

−4x = 8

Step 4: Divide both sides by -4.

−4x/−4

= 8/−4

x = −2

katrin2010 [14]2 years ago

3 0

Answer:

x=-2

Step-by-step explanation:

first distribute the 2 through x-4 you get 2x-8=6x then subtract 2x from both sides of the equation now you have 4x=-8 then divide 4 from both sides of the equation and you x=-2

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Use Lagrange multipliers to find the maximum and minimum values of (i) f(x,y)-81x^2+y^2 subject to the constraint 4x^2+y^2=9. (i

sp2606 [1]

i. The Lagrangian is

$L(x,y,\lambda)=81x^2+y^2+\lambda(4x^2+y^2-9)$

with critical points whenever

$L_x=162x+8\lambda x=0\implies2x(81+4\lambda)=0\implies x=0\text{ or }\lambda=-\dfrac{81}4$

$L_y=2y+2\lambda y=0\implies2y(1+\lambda)=0\implies y=0\text{ or }\lambda=-1$

$L_\lambda=4x^2+y^2-9=0$

If $x=0$ , then $L_\lambda=0\implies y=\pm3$ .
If $y=0$ , then $L_\lambda=0\implies x=\pm\dfrac32$ .
Either value of $\lambda$ found above requires that either $x=0$ or $y=0$ , so we get the same critical points as in the previous two cases.

We have $f(0,-3)=9$ , $f(0,3)=9$ , $f\left(-\dfrac32,0\right)=\dfrac{729}4=182.25$ , and $f\left(\dfrac32,0\right)=\dfrac{729}4$ , so $f$ has a minimum value of 9 and a maximum value of 182.25.

ii. The Lagrangian is

$L(x,y,z,\lambda)=y^2-10z+\lambda(x^2+y^2+z^2-36)$

with critical points whenever

$L_x=2\lambda x=0\implies x=0$ (because we assume $\lambda\neq0$ )

$L_y=2y+2\lambda y=0\implies 2y(1+\lambda)=0\implies y=0\text{ or }\lambda=-1$

$L_z=-10+2\lambda z=0\implies z=\dfrac5\lambda$

$L_\lambda=x^2+y^2+z^2-36=0$

If $x=y=0$ , then $L_\lambda=0\implies z=\pm6$ .
If $\lambda=-1$ , then $z=-5$ , and with $x=0$ we have $L_\lambda=0\implies y=\pm\sqrt{11}$ .

We have $f(0,0,-6)=60$ , $f(0,0,6)=-60$ , $f(0,-\sqrt{11},-5)=61$ , and $f(0,\sqrt{11},-5)=61$ . So $f$ has a maximum value of 61 and a minimum value of -60.

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prisoha [69]

7x + 3 is the final answer

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