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3 years ago

Differentiate between a fixed and movable pulley.

Mathematics

1 answer:

Tamiku [17]3 years ago

8 0

Answer:

Step-by-step explanation:

A fixed pulley consists of a wheel fixed to a shaft and is used in conjunction with a belt to transfer energy to another fixed pulley. A movable pulley consists of a shell, a movable wheel and a rope. Movable pulleys are also know as block and tackle.

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Un padre tiene actualmente cuatro veces la edad de su hijo. Cuando pasen 5 años, su edad será solo tres veces superior. ¿Qué eda

julsineya [31]

Answer:

I. Hijo, h = 10 años

II. Padre, P = 40 años

Step-by-step explanation:

Sea la edad del padre P.
Sea la edad del hijo h.

Traduciendo el problema verbal a una expresión algebraica, tenemos;

P = 4h .....ecuación 1.

P + 5 = 3(h + 5) ........ecuación 2.

Simplificando aún más, tenemos;

P + 5 = 3h + 15

P = 3h + 15 - 5

F = 3h + 10 ......ecuación 3.

Sustituyendo la ecuación 1 en la ecuación 3;

4h = 3h + 10

4h - 3h = 10

h = 10 años

A continuación, encontraríamos la edad del padre;

P = 4h

P = 4 * 10

P = 40 años

4 0

3 years ago

Help?? Me pleaseeee, you must help

storchak [24]

C is your answer e cause u supposed to multiple

7 0

4 years ago

Read 2 more answers

If anyone knows about definite integrals for calculus then please I request help! I

kicyunya [14]

Answer:

$\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{1}{8} \bigg( e^\Big{\frac{4}{25}} - e^\Big{\frac{4}{81}} \bigg)$

General Formulas and Concepts:

Calculus

Differentiation

Derivatives
Derivative Notation

Derivative Property [Multiplied Constant]: $\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)$

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Integration

Integrals

Integration Rule [Fundamental Theorem of Calculus 1]: $\displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)$

Integration Property [Multiplied Constant]: $\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx$

U-Substitution

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx$

Step 2: Integrate Pt. 1

Identify variables for u-substitution.

Set u: $\displaystyle u = 4x^{-2}$
[u] Differentiate [Basic Power Rule, Derivative Properties]: $\displaystyle du = \frac{-8}{x^3} \ dx$
[Bounds] Switch: $\displaystyle \left \{ {{x = 9 ,\ u = 4(9)^{-2} = \frac{4}{81}} \atop {x = 5 ,\ u = 4(5)^{-2} = \frac{4}{25}}} \right.$

Step 3: Integrate Pt. 2

[Integral] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8}\int\limits^9_5 {\frac{-8}{x^3}e^\big{4x^{-2}}} \, dx$
[Integral] U-Substitution: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8}\int\limits^{\frac{4}{81}}_{\frac{4}{25}} {e^\big{u}} \, du$
[Integral] Exponential Integration: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8}(e^\big{u}) \bigg| \limits^{\frac{4}{81}}_{\frac{4}{25}}$
Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8} \bigg( e^\Big{\frac{4}{81}} - e^\Big{\frac{4}{25}} \bigg)$
Simplify: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{1}{8} \bigg( e^\Big{\frac{4}{25}} - e^\Big{\frac{4}{81}} \bigg)$