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3 years ago

Help me answer these questions

Mathematics

1 answer:

Reil [10]3 years ago

8 0

Answer:

See below.

Step-by-step explanation:

1. x = e^(x/y) Taking logs:

log x = x/y

x = y log x

Differentiating:

1 = dy/dx log x + y * 1/x

dy/dx log x = (1 - y/x)

dy/dx = (1 - y/x) / log x

dy/dx = ((x - y)/ x) / log x

dy/dx = (x - y) / x log x)

2. y^x = e ^(y - x)

Taking logs:

x log y = y - x --------------(A)

y = x + x logy --------------(B)

dy/dx = 1 + x * 1/y * dy/dx + 1*logy

dy/dx - dy/dx * x/y = 1 + log y

dy/dx ( (x - y) y)) = 1 + log y

dy/dx = y(1 + log y) / (y - x)

Using (A) and (B) :-

dy/dx = x(1 + logy)(1 + logy) / x logy The x's cancel, so:

dy/dx = (1 + logy)^2 / log y

Sorry I have to go right now..

The rest of the answers are on the re-post of these questions

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Suppose a force of 30 N is required to stretch and hold a spring 0.1 m from its equilibrium position. a. Assuming the spring obe

AlexFokin [52]

Answer:

a) $k = 300\,\frac{N}{m}$ , b) $\Delta U_{k} = 13.5\,J$ , c) $\Delta U_{k} = 6\,J$ , d) $\Delta U_{k} = 4.5\,J$

Step-by-step explanation:

a) The spring constant is calculated by using this expression:

$k = \frac{F}{x}$

$k = \frac{30\,N}{0.1\,m}$

$k = 300\,\frac{N}{m}$

b) The work needed to compress the spring from its initial position is:

$\Delta U_{k} = \frac{1}{2}\cdot k \cdot (x_{f}^{2}-x_{o}^{2})$

$\Delta U_{k} = \frac{1}{2}\cdot (300\,\frac{N}{m} )\cdot [(-0.3\,m)^{2}-(0\,m)^{2}]$

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c) The work needed to stretch the spring is:

$\Delta U_{k} = \frac{1}{2}\cdot (300\,\frac{N}{m} )\cdot [(0.2\,m)^{2}-(0\,m)^{2}]$

$\Delta U_{k} = 6\,J$

d) The work need to stretch the spring is:

$\Delta U_{k} = \frac{1}{2}\cdot (300\,\frac{N}{m} )\cdot [(0.2\,m)^{2}-(0.1\,m)^{2}]$

$\Delta U_{k} = 4.5\,J$

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4 years ago

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Which values are in the solution set of the compound inequality? Check all that apply. 4(x + 3) ≤ 0 or x + 1 > 3 –6 –3 0 3 8

Colt1911 [192]

we have

$4(x + 3) \leq 0$ -------> inequality 1

$x + 1 > 3$ -------> inequality 2

we know that

In this system of inequalities, for a value to be the solution of the system, it is enough that it satisfies at least one of the two inequalities.

let's check each of the values

case 1) x=-6

Substitute the value of x=-6 in the inequality 1

$4(-6 + 3) \leq 0$

$4(-3) \leq 0$

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The value of x=-6 is a solution of the compound inequality-----> It is not necessary to check the second inequality, because the first one satisfies

case 2) x=-3

Substitute the value of x=-3 in the inequality 1

$4(-3 + 3) \leq 0$

$4(0) \leq 0$

$0 \leq 0$ -------> is ok

The value of x=-3 is a solution of the compound inequality-----> It is not necessary to check the second inequality, because the first one satisfies

case 3) x=0

Substitute the value of x=0 in the inequality 1

$4(0 + 3) \leq 0$

$4(3) \leq 0$

$12 \leq 0$ -------> is not ok

Substitute the value of x=0 in the inequality 2

$0 + 1 > 3$

$1 > 3$ --------> is not ok

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case 4) x=3

Substitute the value of x=3 in the inequality 1

$4(3 + 3) \leq 0$

$4(6) \leq 0$

$24 \leq 0$ -------> is not ok

Substitute the value of x=3 in the inequality 2

$3 + 1 > 3$

$4 > 3$ --------> is ok

The value of x=3 is a solution of the compound inequality

case 5) x=8

Substitute the value of x=8 in the inequality 1

$4(8 + 3) \leq 0$

$4(11) \leq 0$

$44 \leq 0$ -------> is not ok

Substitute the value of x=8 in the inequality 2

$8 + 1 > 3$

$9 > 3$ --------> is ok

The value of x=8 is a solution of the compound inequality

case 6) x=10

Substitute the value of x=10 in the inequality 1

$4(10 + 3) \leq 0$

$4(13) \leq 0$

$52 \leq 0$ -------> is not ok

Substitute the value of x=10 in the inequality 2

$10+ 1 > 3$

$11 > 3$ --------> is ok

The value of x=10 is a solution of the compound inequality

therefore

the answer is

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