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mina [271]
3 years ago
8

Can someone explain this​

Mathematics
1 answer:
hodyreva [135]3 years ago
3 0

Answer:

x² = V19²-10²

= V361-100

= V261

x = 16.2

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Explain how you know if an equation has no solutions, one solution or an infinite number of solutions
topjm [15]

Answer:

If you cancel out all of the x terms via addition or subtraction, and you get something along the lines of 1 = 2, then you have no solution.

11x + 4 = 11x + 7 Subtract 11x from both sides

4 = 7 No value for x will satisfy this equation.

If you cancel out all the x terms via addition or subtraction and you get something along the lines of 1 = 1, then you have infinite solutions.

2(x + 1) = 2x + 2 Expand the left side using the distributive property

2x + 2 = 2x + 2 Subtract 2x from both sides

2 = 2 Every value for x will satisfy this equation

If you can’t cancel out all the x terms with addition or subtraction, you probably have 1 solution.

5x + 2 = 3x + 100 Subtract 3x from both sides

2x + 2 = 100 Subtract 2 from both sides

2x = 98 Divide by 2 on both sides

x = 49 The only x value that satisfies this equation is 49

There are other cases where functions of x aren’t injective, meaning there’s more than one x value that satisfies the equation. Here’s what I mean.

x^2 = 4

x = 2 or -2

x^3 = 1

x = 1, -1/2 + isqrt(3)/2, -1/2 - isqrt(3)/2

sin(x) = 0

x = 2n*pi, where n is some integer

5 0
3 years ago
Could you please help me out ?
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What is 207.4÷61pls help
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3.4 is the answer (rounded up of course)
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Which statement about rectangles and rhombuses is always true
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This extreme value problem has a solution with both a maximum value and a minimum value. Use Lagrange multipliers to find the ex
olga nikolaevna [1]

Answer:

maximum value , f max = 169 (taking the most out of the 4th power)

minimum value , f min = 169/3 (taking the least out of the 4th power)

Step-by-step explanation:

A) Since both function and restriction are symmetrical with respect to x,y and z, there is no reason for one to be more important than the others and therefore one solution would be x=y=z=λ and thus

x2 + y2 + z2 = 13  → 3λ² = 13 →λ² = 13/3

and f would be

f (x, y, z) = x4 + y4 + z4 = 3λ⁴ = 3*(13/3)²=13²/3=169/3

since x⁴ increases faster than 3*x² , f(x,y,z) would be a minimum

and the maximum value would be obtained taking the most out of x⁴, thus doing 2 coordinates =0 ( can be x=0 and y=0) and

z²= 13

f (x, y, z) = x4 + y4 + z4 = 13² = 169

B) strictly, using Lagrange multipliers

f (x, y, z) = x4 + y4 + z4

g (x, y, z) = x2 + y2 + z2 - 13

F(x,y,z) = f (x, y, z) -λ*g (x, y, z)

such that

Fx (x,y,z)=  fx(x, y, z) -λ*gx (x, y, z) = 0 → 4*x³ - λ*2*x = 0 → 2*x*(2*x² -λ) = 0

thus x=0 or x²= λ/2

Fy (x,y,z)=  fy(x, y, z) -λ*gy (x, y, z)= 0 → 4*y³ - λ*2*y = 0 → 2*y*(2*y² - λ) = 0

thus y=0 or y²= λ/2

Fz (x,y,z)=  fz(x, y, z) -λ*gz (x, y, z)= 0 → 4*z³ - λ*2*z = 0→ 2*z*(2*z² - λ) = 0

thus z=0 or z²= λ/2

g (x, y, z) = 0  → x2 + y2 + z2 = 13 → 3*(λ/2) = 13 → λ=13*2/3

thus  x²=y²=z²= λ/2 =13/3

f min = f (x, y, z) = x4 + y4 + z4 = 3*(13/3)²=169/3

for the x=0 , y=0 → z²= 13

f max = f (x, y, z) = x4 + y4 + z4 = 13² = 169

3 0
3 years ago
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