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kompoz [17]
3 years ago
15

A) ¿Qué nombre recibe la tabla?B) Elabore una distribución de frecuencias acumulativas y represente la distribución en un polí-g

ono de frecuencias acumulativas.C) De acuerdo con el polígono de frecuencias acumulativas, ¿cuántos empleados ganan $11.00 o menos la hora? ¿La mitad de los empleados ganan más? ¿Cuatro empleados ganan menos?
Mathematics
1 answer:
Alex Ar [27]3 years ago
7 0

Ok, primero haces estas hostilidades de considerable duración y magnitud. En el uso de las ciencias sociales, se agregan ciertas calificaciones. Los sociólogos suelen aplicar el término a tales conflictos sólo si se inician y conducen de acuerdo con formas socialmente reconocidas. Tratan la guerra como una institución reconocida en la costumbre o en la ley. Los escritores militares suelen limitar el término a las hostilidades en las que los grupos contendientes tienen el mismo poder para hacer que el resultado sea incierto durante un tiempo. Los conflictos armados de estados poderosos con pueblos aislados e impotentes suelen denominarse pacificaciones, expediciones militares o exploraciones; con los estados pequeños, se les llama intervenciones o represalias; y con grupos internos, rebeliones o insurrecciones. Tales incidentes, si la resistencia es lo suficientemente fuerte o prolongada, pueden alcanzar una magnitud que les dé derecho al nombre de "guerra".

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Write the following expression using exponents.<br> 4x4x4x4 =
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Answer:

4^4

Step-by-step explanation:

Since there are four 4's, it is to the 4th power.

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1. (5pts) Find the derivatives of the function using the definition of derivative.
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2.8.1

f(x) = \dfrac4{\sqrt{3-x}}

By definition of the derivative,

f'(x) = \displaystyle \lim_{h\to0} \frac{f(x+h)-f(x)}{h}

We have

f(x+h) = \dfrac4{\sqrt{3-(x+h)}}

and

f(x+h)-f(x) = \dfrac4{\sqrt{3-(x+h)}} - \dfrac4{\sqrt{3-x}}

Combine these fractions into one with a common denominator:

f(x+h)-f(x) = \dfrac{4\sqrt{3-x} - 4\sqrt{3-(x+h)}}{\sqrt{3-x}\sqrt{3-(x+h)}}

Rationalize the numerator by multiplying uniformly by the conjugate of the numerator, and simplify the result:

f(x+h) - f(x) = \dfrac{\left(4\sqrt{3-x} - 4\sqrt{3-(x+h)}\right)\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{\left(4\sqrt{3-x}\right)^2 - \left(4\sqrt{3-(x+h)}\right)^2}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{16(3-x) - 16(3-(x+h))}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ f(x+h) - f(x) = \dfrac{16h}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)}

Now divide this by <em>h</em> and take the limit as <em>h</em> approaches 0 :

\dfrac{f(x+h)-f(x)}h = \dfrac{16}{\sqrt{3-x}\sqrt{3-(x+h)}\left(4\sqrt{3-x} + 4\sqrt{3-(x+h)}\right)} \\\\ \displaystyle \lim_{h\to0}\frac{f(x+h)-f(x)}h = \dfrac{16}{\sqrt{3-x}\sqrt{3-x}\left(4\sqrt{3-x} + 4\sqrt{3-x}\right)} \\\\ \implies f'(x) = \dfrac{16}{4\left(\sqrt{3-x}\right)^3} = \boxed{\dfrac4{(3-x)^{3/2}}}

3.1.1.

f(x) = 4x^5 - \dfrac1{4x^2} + \sqrt[3]{x} - \pi^2 + 10e^3

Differentiate one term at a time:

• power rule

\left(4x^5\right)' = 4\left(x^5\right)' = 4\cdot5x^4 = 20x^4

\left(\dfrac1{4x^2}\right)' = \dfrac14\left(x^{-2}\right)' = \dfrac14\cdot-2x^{-3} = -\dfrac1{2x^3}

\left(\sqrt[3]{x}\right)' = \left(x^{1/3}\right)' = \dfrac13 x^{-2/3} = \dfrac1{3x^{2/3}}

The last two terms are constant, so their derivatives are both zero.

So you end up with

f'(x) = \boxed{20x^4 + \dfrac1{2x^3} + \dfrac1{3x^{2/3}}}

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What two letters don't appear on the telephone dial​
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After a dilation with center (0,0) the image of DB is D’B’ if DB=6 and D’B’=2 the scale factor of dilation is what ?
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scale factor = \frac{1}{3}

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the scale factor is the ratio of corrresponding sides. image to original.

scale factor = \frac{D'B'}{DB} = \frac{2}{6} = \frac{1}{3}

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