Find the nth term -2 ,1,4,7

Mathematics

1 answer:

Ira Lisetskai [31]3 years ago

6 0

-2, 1, 4, 7, 10, 13, 16, 19, 22

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Mumz [18]

Answer:

The answer is 4 zeros

Step-by-step explanation:

(6×5)×10³

30×1000=30000

Therefore 4 zeros in 30000

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3 years ago

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Find the vertical asymptote(s) of f of x equals quantity 3 x squared plus 3x plus 6 end quantity over quantity x squared minus 2

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Geometry question shown above. Please help.

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For question 4 its (C)

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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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2 years ago

What is the value of X?

dlinn [17]

Answer

The value of x is 180 degrees

Explanation

you add 150 and 30 to get the value of X plus if you look carefully you can see that in the image x is at 180 degrees.

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2 years ago