What is the minimum value for g(x)=x2−10x+16 ?

Mathematics

1 answer:

mr_godi [17]2 years ago

3 0

G(x)=x^2-10x+16
g(x)=x^2-10x+25-25+16
g(x)= (x-5)^2-9
minimum value is -9

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Please show work! Thank you

xz_007 [3.2K]

A) The greatest rectangular area will be the area of a square 10 m on each side, 100 m^2.

b) The new dimensions will be 11 m × 11 m.
.. The new area will be (11 m)^2 = 121 m^2.

c) The area was increased by 121 m^2 -100 m^2 = 21 m^2, or 21%.

d) Yes, and no.
.. If you increase the dimensions by 10%, the area will increase by 21%.
.. (40 m)^2 = 1600 m^2
.. (44 m)^2 = 1936 m^2 = 1.21*(1600 m^2), an increase of 21% over the original.

.. If you increase the dimensions by 1 unit, the area will increase by (2x+1) square units, where x is the side of the original. For x≠10, this is not 21 square units.
.. (41 m)^2 = 1681 m^2 = 1600 m^2 +(2*40 +1) m^2 = 1600 m^2 +81 m^2

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Answer:

Step-by-step explanation:

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2 years ago

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Determine the quadrant when the terminal side of the angle lies according to the following conditions: cos (t) < 0, csc (t) &

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Answer:

The angle is in the second quadrant.

Step-by-step explanation:

The cosecant of an angle is the same as the reciprocal of the sine of that angle. In other words, as long as $\sin (t) \ne 0$ ,

$\displaystyle \csc t = \frac{1}{\sin t}$ .

Therefore, $\csc(t) > 0$ is equivalent to $\sin (t) > 0$ .

Consider a unit circle centered at the origin. If the terminal side of angle $t$ intersects the unit circle at point $(x,\, y)$ , then

$\cos (t) = x$ , and
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For angle $t$ ,

$x = \cos(t) < 0$ , meaning that the intersection is to the left of the $y$ -axis.
$y = \sin(t) > 0$ , meaning that the intersection is above the $x$ -axis.