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2 years ago

The weight Wkg of a metal bar varies jointly

Mathematics

1 answer:

maria [59]2 years ago

6 0

Answer:

d = $\sqrt{\frac{216W}{35L} }$

Step-by-step explanation:

Given that W varies jointly as L and d² then the equation relating them is

W = kLd² ← k is the constant of variation

To find k use the condition W = 140 when d = 4 and L = 54, thus

140 = k × 54 × 4² = 864k ( divide both sides by 864 )

$\frac{140}{864}$ = k , that is

k = $\frac{35}{216}$

W = $\frac{35}{216}$ Ld² ← equation of variation

Multiply both sides by 216

216W = 35Ld² ( divide both sides by 35L )

$\frac{216W}{35L}$ = d² ( take the square root of both sides )

d = $\sqrt{\frac{216W}{35L} }$

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7 0

2 years ago

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Serga [27]

Answer:

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3 years ago

Find the minimum and maximum value of the function on the given interval by comparing values at the critical points and endpoint

Kitty [74]

Answer:

maximum: y = 1

minimum: y = 0.

Step-by-step explanation:

Here we have the function:

y = f(x) = √(1 + x^2 - 2x)

we want to find the minimum and maximum in the segment [0, 1]

First, we evaluate in the endpoints, which are 0 and 1.

f(0) =√(1 + 0^2 - 2*0) = 1

f(1) = √(1 + 1^2 - 2*1) = 0

Now let's look at the critical points (the zeros of the first derivate)

To derivate our function, we can use the chain rule:

f(x) = h(g(x))

then

f'(x) = h'(g(x))*g(x)

Here we can define:

h(x) = √x

g(x) = 1 + x^2 - 2x

Then:

f(x) = h(g(x))

f'(x) = 1/2*( 1 + x^2 - 2x)*(2x - 2)

f'(x) = (1 + x^2 - 2x)*(x - 1)

f'(x) = x^3 - 3x^2 + x - 1

this function does not have any zero in the segment [0, 1] (you can look it in the image below)

Thus, the function does not have critical points in the segment.

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3 years ago

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sdas [7]

Answer:

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