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alexandr402 [8]
3 years ago
11

Graph the line passing through (−2,5) whose slope is m=1/5

Mathematics
2 answers:
horrorfan [7]3 years ago
8 0

Answer:

In order to graph the line, you need to first find the y-intercept (aka 'b').

Equation: y=mx+b

Given the two points and slope, we can plug in and find for b

5 = 1/5(-2) + b

5 = -2/5 + b

b = 5 + 2/5

b = 27/5 or 5.4

So your final equation will be: y=1/5x+27/5

Lastly you just need to plot the 'b = 5.4'  on the y-axis and place the slope as 1 over 5.

Nonamiya [84]3 years ago
4 0

Hello there! :)

Given:

x-coordinate of point = -2

y-coordinate of point = 5

Slope, or "m" = 1/5

Substitute in these values into slope-intercept formula (y = mx+b)

5 = 1/5(-2) + b

5 = -2/5 + b

25/5 = -2/5 + b

27/5 = b

Rewrite the equation:

y = 1/5x + 27/5

Graph the equation using a graphing utility:

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Answer:

JL = 52

Step-by-step explanation:

It is given in the question that K is the midpoint of JL.

So, JK = KL ......... (1)

Now, given that, JK =3x - 1 and KL = 2x + 8  

Therefore, from equation (1), we can write  

3x - 1 = 2x + 8

⇒ x = 9

So, JK = 3x - 1 = 3(9) -1 = 26 and KL = 2x + 8 = 2(9) + 8 = 26

Hence, JL = JK + KL = 26 + 26 = 52. (Answer)

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You ran 800 meters in 5 minutes what your average speed
larisa [96]

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160 metres per minute

Step-by-step explanation:

800 / 5 = 160

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Which property of equality would you use first to solve the equation 6x = 3x
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Step-by-step explanation:

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3 years ago
Can someone help me
erik [133]

Answer:

   6 < x < 23.206

Step-by-step explanation:

To properly answer this question, we need to make the assumption that angle DAC is non-negative and that angle BCA is acute.

The maximum value of the angle DAC can be shown to occur when points B, C, and D are on a circle centered at A*. When that is the case, the sine of half of angle DAC is equal to 16/22 times the sine of half of angle BAC. That is, ...

  (2x -12)/2 = arcsin(16/22×sin(24°))

  x ≈ 23.206°

Of course, the minimum value of angle DAC is 0°, so the minimum value of x is ...

  2x -12 = 0

  x -6 = 0 . . . . . divide by 2

  x = 6 . . . . . . . add 6

Then the range of values of x will be ...

  6 < x < 23.206

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* One way to do this is to make use of the law of cosines:

  22² = AB² + AC² -2·AB·AC·cos(48°)

  16² = AD² + AC² -2·AD·AC·cos(2x-12)

The trick is to maximize x while satisfying the constraints that all of the lengths are positive. This will happen when AB=AC=AD, in which case the equations be come ...

  22² = 2·AB²·(1-cos(48°))

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The value of AB drops out of the ratio of these equations, and the result for x is as above.

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3 years ago
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WILL MARK BRAINLIEST!!! PLEASE ANSWER!!! WRONG ANSWERS JUST FOR POINTS WILL BE REPORTED!!!!!!!!!!!!!!
alexgriva [62]

Answer:

916.1 cm^3

Step-by-step explanation:

1) find the radius of the cylinder:

radius = dimater / 2 = 7,3 / 2 = 3.65 cm

2) find the base area of the cylinder:

base area = radius^2 x 3.14 = 3.65^2 x 3.14 = 41.83265 cm^2

3) find the height of the cylinder:

height = 7.3 x 3 = 21.9 cm

4) find the volume

V = base area x height = 41.83265 x 21.9 = 916.135035 cm^3

5) round to nearest tenth:

916. 1 cm^3

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