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Alex17521 [72]
3 years ago
6

Number 2 please help

Mathematics
1 answer:
Sveta_85 [38]3 years ago
5 0

Answer:

you just have to check the first and the third box

Step-by-step explanation:

put each into slop intercept form

y+6=-5(x-2) is the same as y=-5x+4

y-2=-5(x+6) is the same as y=-5x-28

y=-5x+4

y=-5x-28

all of them have a slope of -5, but you need to plug in (2,-6) to see which on is true

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vovikov84 [41]
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Solve the separable differential equation dtdx=x2+164 and find the particular solution satisfying the initial condition x(0)=9
maria [59]
\dfrac{dt}{dx} = x^2 + \frac{1}{64} \Rightarrow\ dt = \left(x^2 + \frac{1}{64}\right)dx \Rightarrow \\ \\ \displaystyle \int 1 dt = \int \left(x^2 + \frac{1}{64}\right)dx \Rightarrow \\ t = \dfrac{x^3}{3} + \frac{x}{64} + C

C = 9 because all the x terms go away.

t = \dfrac{x^3}{3} + \dfrac{x}{64} + 9
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3 years ago
Find an equation in slope-intercept form for the line with slope -12 and y-intercept = 1/2
Kruka [31]
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Find the slope-intercept equation
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4 0
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Step-by-step explanation:

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Which of the following operations is true regarding relative frequency distributions? Multiple choice question. No two classes c
pshichka [43]

Answer:

The relative frequency is found by dividing the class frequencies by the total number of observations

Step-by-step explanation:

Relative frequency measures how often a value appears relative to the sum of the total values.

An example of how relative frequency is calculated

Here are the scores and frequency of students in a maths test

Scores (classes)              Frequency                Relative frequency

0 - 20                                10                               10 / 50 = 0.2

21 - 40                               15                               15 / 50 = 0.3

41 - 60                               10                               10 / 50 = 0.2

61 - 80                                5                                 5 / 50  = 0.1

81 - 100                             <u> 10</u>                                10 / 50 = <u>0.2</u>

                                          50                                               1

From the above example, it can be seen that :

  1. two or more classes  can have the same relative frequency
  2. The relative frequency is found by dividing the class frequencies by the total number of observations.
  3. The sum of the relative frequencies must be equal to one
  4. The sum of the frequencies and not the relative frequencies is equal to the number of observations.

4 0
3 years ago
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