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alexandr1967 [171]
3 years ago
12

“Draw the graph of the line with equation y=3/4x-2”

Mathematics
1 answer:
Ugo [173]3 years ago
3 0

Step-by-step explanation:

You have multiple ways, all revolving around finding 2 points.

The easiest point to use is the y intercept (the value of y when x=0), and is the constant term, giving you (0; -2)

The second point can be found in a number of ways:

  • Look for the x intercept (when y=0), and you will get some point by solving for x;
  • put any value of x you haven't used so far (ie, 0), compute it, and you will find a second point.
  • plug a "smarter" value of x that will make computation easy: in this case you have a fraction, so I would pick the denominator to simplify: (you got \frac34, so x=4 \rightarrow y=\frac34(4)-2 = 3-2=1 giving a second point (4;1))
  • since the equation is in <em>slope-intercept </em>form, Just remember the slope tells you how many vertical units you move (numerator) when you move right the number of units in the denominator, with the sign telling you up or down: in our case you move 3 up (since it's positive) every 4 "steps" right.

In any case, the graph will look something like the one in the image

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Your answer is A. The sum of x and y is a rational number .

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Suppose an arrow is shot upward on the moon with a velocity of 44 m/s, then its height in meters after tt seconds is given by h(
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Answer:

a. 38.19m/s

b. 38.605m/s

c. 38.937m/s

d. 39.0117m/s

e. 39.01917m/s

Step-by-step explanation:

The average velocity is defined as the relationship between the displacement that a body made and the total time it took to perform it. Mathematically is given by the next formula:

v_a_v_g = \frac{\Delta x}{\Delta t} =\frac{x_f-x_i}{t_f-t_i}

Where:

x_f=Final\hspace{3}distance\hspace{3}traveled\\x_i=Initial\hspace{3}distance\hspace{3}traveled\\t_f=Final\hspace{3}time\hspace{3}interval\\t_i=Initial\hspace{3}time\hspace{3}interval

a. Let's find h(3) and h(4) using the data provided by the problem:

h(3)=44(3)-0.83(3^2)=124.53=x_i\\h(4)=44(4)-0.83(4^2)=162.72=x_f

The average velocity over the interval [3, 4] is :

v_a_v_g=\frac{162.72-124.53}{4-3} =38.19m/s

b. Let's find h(3.5) using the data provided by the problem:

h(3.5)=44(3.5)-0.83(3.5^2)=143.8325=x_f

The average velocity over the interval [3, 3.5] is :

v_a_v_g=\frac{143.8325-124.53}{3.5-3} =38.605m/s

c. Let's find h(3.1) using the data provided by the problem:

h(3.1)=44(3.1)-0.83(3.1^2)=128.4237=x_f

The average velocity over the interval [3, 3.1] is :

v_a_v_g=\frac{128.4237-124.53}{3.1-3} =38.937m/s

d. Let's find h(3.01) using the data provided by the problem:

h(3.1)=44(3.01)-0.83(3.01^2)=124.920117=x_f

The average velocity over the interval [3, 3.01] is :

v_a_v_g=\frac{124.920117-124.53}{3.01-3} =39.0117m/s

e. Let's find h(3.001) using the data provided by the problem:

h(3.001)=44(3.001)-0.83(3.001^2)=124.5690192=x_f

v_a_v_g=\frac{124.5690192-124.53}{3.001-3} =39.01917m/s

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Answer:

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then to check it you can multiply

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Step-by-step explanation:

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