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3 years ago

What is the equation of the quadratic graph with a focus of (5, 6) and a directrory of y = 2?

1 answer:

4 0

The vertex of the graph is at (5, (6 + 2)/2) = (5, 4)
The equation of a quadratic graph is given by y - k = 4p(x - h)^2, where (h, k) is the vertex, p is the distance from the vertex to the focus.
Here, (h, k) = (5, 4) and p = 6 - 2 = 2 and since the focus is on top of the directrix, the parabola is facing up and the value of p is positive.
Therefore, the required equation is y - 4 = 4(2)(x - 5)^2
y - 4 = 8(x^2 - 10x + 25)
y - 4 = 8x^2 - 80x + 200
y = 8x^2 - 80x + 204

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A spyware is trying to break into a system by guessing its password. It does not give up until it tries 1 million different pass

Anettt [7]

Answer:

$Probability = 0.006033$

Step-by-step explanation:

Given

$Tries = 1000000$

Required

Determine the probability of 6 different lower case letters (Question continuation)

There are 26 lower case letters.

The first can be any of letters 26

The second can be any of letters 26 - 1

The third can be any of letters 26 - 2

The fourth can be any of letters 26 - 3

The fifth can be any of letters 26 - 4

The sixth can be any of letters 26 - 5

Number of selection is:

$Selection = 26 * (26 - 1) * (26 - 2) * (26 - 3) * (26 - 4) * (26 - 5)$

$Selection = 26 * 25 * 24 * 23 * 22 * 21$

$Selection = 165765600$

The probability is:

$Probability = \frac{Tries}{Selection}$

$Probability = \frac{1000000}{165765600}$

$Probability = 0.00603261472$

$Probability = 0.006033$ --- approximated

6 0

2 years ago

I need help please thank you

aliya0001 [1]

$\frac{ x^{2}-25 }{ x^{2}-3x-10 }$
$\frac{(x+5)(x-5)}{(x-5)(x+2)}$
$\frac{x-5}{x+2}$

5 0

3 years ago

Please help I don’t understand at all if you could possibly break it down that would be nice.

Yuri [45]

sub 2 for any "x"

3(2)^2-2(2)+1

6^2-4+1 i dont know if you need to simplify even more but if you do

36-4+1 32+1 31

i hope this helps you!

4 0

3 years ago

Read 2 more answers

Find the fifth term of the geometric sequence whose first term is 81 and whose common ratio is 1/3.(1 point)13

Ludmilka [50]

$n^{th}\textit{ term of a geometric sequence} \\\\ a_n=a_1\cdot r^{n-1}\qquad \begin{cases} a_n=n^{th}\ term\\ n=\stackrel{\textit{term position}}{5}\\ a_1=\stackrel{\textit{first term}}{81}\\ r=\stackrel{\textit{common ratio}}{\frac{1}{3}} \end{cases}\implies a_5=81\left( \cfrac{1}{3} \right)^{5-1} \\\\\\ a_5=81\left( \cfrac{1}{3} \right)^4\implies a_5=81\cdot \cfrac{1}{81}\implies a=1$

7 0

2 years ago

Find the 31st term of the following sequence. 9, 15, 21, ...

Svet_ta [14]

Notice that the above sequence is arithmetic because the common difference: d is constant.

d = second term - first term.

= 15 - 9

= 6

The general term of an arithmetic sequence is,

a_{n} =a_{1} +(n-1)d

Where , nth term = $a_{n}$

$a_{1}$ = First term

and d = common difference.

The given sequence is: 9, 15, 21, ...

Here a_{1} = 9,

d = 6

We need to find the 31st term. So, n = 31.

Next step is to plug in these values in the above formula. Therefore,

$a_{31} =9 +(31-1)*6$

= 9 + 30* 6

= 9 + 180

= 189

So, 31st term of this sequence is 189.

Hope this helps you.

4 0

3 years ago