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Vesna [10]
3 years ago
10

Daniel hiked 4 5/8 miles on Saturday and 4 1/2 miles on Sunday. How many miles did he hike in all?

Mathematics
1 answer:
Ivenika [448]3 years ago
7 0

Answer:

He hiked 9 1/8 miles

Step-by-step explanation:

First make the fractions like terms

Multiply 4 1/2 by 4/4 to get 4 4/8

Then add the fractions to get 9/8

9/8 is an improper fraction, make it 1 1/8

then add all the whole numbers 4+4+1=9

add the fraction 1/8 to 9= 9 1/8

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Marcus loves baseball and wants to create a home plate for his house. Marcus needs to calculate the area of the home plate at th
Whitepunk [10]

Answer:

For the answer to the question above,  first you must find the area of the rectangle, which is 3 × 6 = 18 inches squared. Then, you find the area of the triangles. To find the area of the top triangle, you'd do base × height/2 (5 × 6)/2 = 15 inches squared. The triangle on the left would be (4 × 3)/2 which is 6 inches squared. After you've added up all of the values, you get (18 + 15 + 6) = 39 inches squared

Step-by-step explanation:

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3 years ago
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How to do that? (Indices)<br><br> a<img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B2%7D%20" id="TexFormula1" title=" \frac
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3 years ago
The function f (x comma y )equals 3 xy has an absolute maximum value and absolute minimum value subject to the constraint 3 x sq
zmey [24]

Answer:

The maximum value of f is 363, which is reached in (11,11) and (-11,-11) and the minimum value of f is -33, which is reached in (√11,-√11) and (-√11,√11)

Step-by-step explanation:

f(x,y) = 3xy, lets find the gradient of f. First lets compute the derivate of f in terms of x, thinking of y like a constant.

f_x(x,y) = 3y

In a similar way

f_y(x,y) = 3x

Thus,

\nabla{f} = (3y,3x)

The restriction is given by g(x,y) = 121, with g(x,y) = 3x²+3y²-5xy. The partial derivates of g are

[ŧex] g_x(x,y) = 6x-5y [/tex]

g_y(x,y) = 6y - 5x

Thus,

\nabla g(x,y) = (6x-5y,6y-5x)

For the Langrange multipliers theorem, we have that for an extreme (x0,y0) with the restriction g(x,y) = 121, we have that for certain λ,

  • f_x(x_0,y_0) = \lambda \, g_x(x0,y0)
  • f_y(x_0,y_0) = \lambda \, g_y(x_0,y_0)
  • g(x_0,y_0) = 121

This can be translated into

  • 3y = \lambda (6x-5y)
  • 3x = \lambda (-5x+6y)
  • 3 (x_0)^2 + 3(y_0)^2 - 5\,x_0y_0 = 121

If we sum the first two expressions, we obtain

3x + 3y = \lambda (x+y)

Thus, x = -y or λ=3.

If x were -y, then we can replace x for -y in both equations

3y = -11 λ y

-3y = 11 λ y, and therefore

y = 0, or λ = -3/11.

Note that y cant take the value 0 because, since x = -y, we have that x = y = y, and g(x,y) = 0. Therefore, equation 3 wouldnt hold.

Now, lets suppose that λ=3, if that is the case, we can replace in the first 2 equations obtaining

  • 3y = 3(6x-5y) = 18x -15y

thus, 18y = 18x

y = x

and also,

  • 3x = 3(6y-5x) = 18y-15x

18x = 18y

x = y

Therefore, x = y or x = -y.

If x = -y:

Lets evaluate g in (-y,y) and try to find y

g(-y,y) = 3(-y)² + 3y*2 - 5(-y)y = 11y² = 121

Therefore,

y² = 121/11 = 11

y = √11 or y = -√11

The candidates to extremes are, as a result (√11,-√11), (-√11, √11). In both cases, f(x,y) = 3 √11 (-√11) = -33

If x = y:

g(y,y) = 3y²+3y²-5y² = y² = 121, then y = 11 or y = -11

In both cases f(11,11) = f(-11,-11) = 363.

We conclude that the maximum value of f is 363, which is reached in (11,11) and (-11,-11) and the minimum value of f is -33, which is reached in (√11,-√11) and (-√11,√11)

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3 years ago
How would you do this?
sdas [7]
You can't:)
destroy the paper
(*´・v・)ε/̵͇̿̿/'̿̿ ̿ ̿ ̿ ̿ ̿
7 0
4 years ago
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