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strojnjashka [21]
3 years ago
7

What is the equation of line m

Mathematics
1 answer:
svlad2 [7]3 years ago
7 0

Answer:

so I dont know what you mean by that

Step-by-step explanation:

there is no way for anyone to tell sorry

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Its the third one down
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What values of x satisfy the equation 2x^2 -3x +4 =0
BaLLatris [955]

Answer:

(3 ± √23 * i) /4

Step-by-step explanation:

To solve this, we can apply the Quadratic Equation.

In an equation of form ax²+bx+c = 0, we can solve for x by applying the Quadratic Equation, or x = (-b ± √(b²-4ac))/(2a)

Matching up values, a is what's multiplied by x², b is what's multiplied by x, and c is the constant, so a = 2, b = -3, and c = 4

Plugging these values into our equation, we get

x = (-b ± √(b²-4ac))/(2a)

x = (-(-3) ± √(3²-4(2)(4)))/(2(2))

= (3 ± √(9-32))/4

= (3 ± √(-23))/4

= (3 ± √23 * i) /4

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3 years ago
Del has 12 less than 3 times the number of comic books as Patrick. Del has at most 57 comic books. What solution represents the
Akimi4 [234]

Answer:

Del has 19 comic book

Step-by-step explanation:

to figure this out we need to take 57 and divide by the 3 given.

57 ÷ 3 = 19

so because we divided 57 by 3 we know the answer is 19

Del has 19 comic books

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2 years ago
This extreme value problem has a solution with both a maximum value and a minimum value. use lagrange multipliers to find the ex
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L(x,y,z,\lambda)=10x+10y+2z+\lambda(5x^2+5y^2+2z^2-42)

L_x=10+10\lambda x=0\implies1+\lambda x=0

L_y=10+10\lambda y=0\implies1+\lambda y=0

L_z=2+4\lambda z=0\implies1+2\lambda z=0

L_\lambda=5x^2+5y^2+2z^2-42=0

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5x^2+5y^2+2z^2=5(2z)^2+5(2z)^2+2z^2=42z^2=42\implies z^2=1

z^2=1\implies z=\pm1\implies x=y=\pm2

There are two critical points, at which we have

f\left(2,2,1\right)=42\text{ (a maximum value)}

f\left(-2,-2,-1\right)=-42\text{ (a minimum value)}

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