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lara [203]
3 years ago
5

La trayectoria de cierto satelitese ajusta ala grafica de la funcionf(x) igual6x al cuadradomenos 12donde x representael tiempo

en días y f(x9 el recorrido en kilometroscuantos kilómetros habrá recorridoel sateliteal cabo de diez días desde su lanzamiento
Mathematics
1 answer:
Nutka1998 [239]3 years ago
5 0

Answer:

588 kilómetros

Step-by-step explanation:

La función con la que estamos trabajando según la pregunta es;

F (x) = 6x ^ 2 -12

Ahora, la pregunta que simplemente nos hace es encontrar el valor de F (x) dado que x = 10

Entonces, lo simple que hacemos aquí es hacer una sustitución de x = 10 Eso sería;

F (10) = 6 (10) ^ 2 - 12 = 600-12 = 588

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maw [93]

Answer:

A, B, C, D, and F are all solutions.

Step-by-step explanation:

All of the possible answers are correct except for E. Answer E is 0.001 less than -4 which would not make the equation true. Hope this helps!!

6 0
3 years ago
36 − 9m^2 factorized
STatiana [176]

Answer:

Step-by-step explanation:

Note that the expression is the difference of squares.

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7 0
2 years ago
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assume that 2000 births are randomly selected and 64 of the births are girls. use subjective judgment to describe the number of
tatuchka [14]

Total no. of births = 2000

Births of girls = 64

Birth% of girls

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=  \frac{64}{20}

=  \frac{6.4}{2}

= 3.2\%

Hence, the number of girls are significantly low.

4 0
3 years ago
How many groups of 2/3 are in 3/5
amid [387]

To find the answer, we will need to divide 3/5 by 2/3.

3/5 / 2/3

3/5 x 3/2

9 / 10

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Hope this helps!! :)

5 0
2 years ago
Derivative, by first principle<br><img src="https://tex.z-dn.net/?f=%20%5Ctan%28%20%5Csqrt%7Bx%20%7D%20%29%20" id="TexFormula1"
vampirchik [111]
\displaystyle\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan x}h

Employ a standard trick used in proving the chain rule:

\dfrac{\tan\sqrt{x+h}-\tan x}{\sqrt{x+h}-\sqrt x}\cdot\dfrac{\sqrt{x+h}-\sqrt x}h

The limit of a product is the product of limits, i.e. we can write

\displaystyle\left(\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan x}{\sqrt{x+h}-\sqrt x}\right)\cdot\left(\lim_{h\to0}\frac{\sqrt{x+h}-\sqrt x}h\right)

The rightmost limit is an exercise in differentiating \sqrt x using the definition, which you probably already know is \dfrac1{2\sqrt x}.

For the leftmost limit, we make a substitution y=\sqrt x. Now, if we make a slight change to x by adding a small number h, this propagates a similar small change in y that we'll call h', so that we can set y+h'=\sqrt{x+h}. Then as h\to0, we see that it's also the case that h'\to0 (since we fix y=\sqrt x). So we can write the remaining limit as

\displaystyle\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan\sqrt x}{\sqrt{x+h}-\sqrt x}=\lim_{h'\to0}\frac{\tan(y+h')-\tan y}{y+h'-y}=\lim_{h'\to0}\frac{\tan(y+h')-\tan y}{h'}

which in turn is the derivative of \tan y, another limit you probably already know how to compute. We'd end up with \sec^2y, or \sec^2\sqrt x.

So we find that

\dfrac{\mathrm d\tan\sqrt x}{\mathrm dx}=\dfrac{\sec^2\sqrt x}{2\sqrt x}
7 0
3 years ago
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