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2 years ago

Number 39 I'm so stuck help plz

Mathematics

1 answer:

stellarik [79]2 years ago

6 0

39 seems very easy. Just add 2 to the number of blocks in square 6. That'll give you Square 7 and so on.

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The functions f(x) = −(x − 1)2 5 and g(x) = (x 2)2 − 3 have been rewritten using the completing-the-square method. apply your kn

ale4655 [162]

The vertex of the function f(x) exists (1, 5), the vertex of the function g(x) exists (-2, -3), and the vertex of the function f(x) exists maximum and the vertex of the function g(x) exists minimum.

<h3>How to determine the vertex for each function is a minimum or a maximum? </h3>

Given:

$\mathrm{f}(\mathrm{x})=-(\mathrm{x}-1)^{2}+5$ and

$\mathrm{g}(\mathrm{x})=(\mathrm{x}-2)^{2}-3$

The generalized equation of a parabola in the vertex form exists

$$y=a(x-h)^{2}+k$

Vertex of the function f(x) exists (1, 5).

Vertex of the function g(x) exists (-2, -3).

Now, if (a > 0) then the vertex of the function exists minimum, and if (a < 0) then the vertex of the function exists maximum.

The vertex of the function f(x) exists at a maximum and the vertex of the function g(x) exists at a minimum.

To learn more about the vertex of the function refer to:

brainly.com/question/11325676

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8 0

1 year ago

How to solve this question

VMariaS [17]

<span>The variance method is as follows.

-Sum the squares of the values in data set, and then divide by the number of values in data set
- From that, subtract the square of the mean (add all values and divide by number of values in the data set)

Our variance is

<span> $\displaystyle\sigma^2 = \frac{2^2 + 5^2 + m^2}{3} - \left(\frac{2 + 5 + m}{3}\right)^2$

Since variance has to be 14, we set $\sigma^2 = 14$ and solve for m

$14= \frac{4 + 25 + m^2}{3} - \left(\frac{7 + m}{3}\right)^2\ \Rightarrow \\ \\
14 = \frac{29}{3} + \frac{1}{3}m^2 - \frac{1}{9}(7+m)^2 \\ \\
14 = \frac{29}{3}+ \frac{1}{3}m^2 - \frac{1}{9}(49 + 14m + m^2) \\ \\
14 = \frac{29}{3}+ \frac{1}{3}m^2 - \frac{49}{9}- \frac{14}{9}m- \frac{1}{9}m^2 \\ \\
0 = \frac{-88}{9} -\frac{14}{9}m + \frac{2}{9}m^2
$

quadratic formula

$m = \displaystyle\frac{-b \pm \sqrt{b^2 -4ac}}{2a} \\
m = \frac{-(-\frac{14}{9}) \pm \sqrt{\left(-\frac{14}{9}\right)^2 - 4(2/9)(-88/9)} }{2(2/9)} \\
m = \frac{\frac{14}{9} \pm \sqrt{ \frac{196}{81} + \frac{704}{81} } }{\frac{4}{9} } \\
m = \frac{\frac{14}{9} \pm \sqrt{ \frac{900}{81} } }{\frac{4}{9} } \\
m = \frac{\frac{14}{9} \pm \sqrt{ \frac{100}{9} } }{\frac{4}{9} } \\
m = \frac{\frac{14}{9} \pm \frac{10}{3} }{\frac{4}{9} } \\
m = 11, -4$

-4 doesnt' work as it is not a positive integer

m = 11

</span></span>

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