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3 years ago

How do I write an equation in point slope form

2 answers:

7 0

The formula is
y-y1 = m(x-x1)
First, find the slope using the points (- 2, 3) and (3, - 1) : m = 3-(-1)/-2-3 = 4/-5 = -4/5
Next, pick a point -- for example, (- 2, 3) . Using this point, x1 = - 2 and y1 = 3 .
Therefore, the equation of this line is y - 3 = -4/5 (x - (- 2)) , which is equivalent to y - 3 = - 4/5(x + 2) .

valentina_108 [34]3 years ago

5 0

Hi there...
the point slope formula is
y-y1=m(x-x1)

where y and x represent any points
y1 and x1 represents known points
and m represents the slope

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What does the magnitude of a vector represent?

nalin [4]

Answer:

A vector's magnitude represents its length, so your answer is C, the length of a vector.

Step-by-step explanation:

7 0

3 years ago

A town has a population of 32,000 in the year 2002; 38,400 in the year 2003; 46,080 in the year 2004; and 55,296 in the year 200

NeTakaya

The point is to find the growth rate. The compound formula is:

P=A(1+ growth rate)ⁿ, where A is the initial Value & P the new value after n years:
P₂₀₀₃ =P₂₀₀₂ (1+ growth rate)¹ (the period "n" from 2002 to 2003 being 1 year)

38400 = 32000(1+growth rate)¹
38400 / 32000 - 1= growth rate & growth rate = 1/5 = 0.2
You will balso find the same growth rate for:
P₂₀₀₄ = P₂₀₀₃(1+ growth rate)¹
P₂₀₀₅ = P₂₀₀₄((1+ growth rate)¹
between 2015 & 2002 THERE ARE 14 YEARS:

P₂₀₁₅ = P₂₀₀₂(1+0.2)¹⁴ & P₂₀₁₅ = 32000(1+02)¹⁴ = 410,854

8 0

3 years ago

Find the quotient: 8 ÷ 0.5

podryga [215]

The quotient is 16. Because 16 0.5s go into 8

6 0

3 years ago

Read 2 more answers

A teacher bought 7 packs of dominoes for a math game. Each pack had 55 dominoes. The teacher already had 118 dominoes to use in

Mamont248 [21]

Answer:

the teacher had 503 dominoes for the game.

Step-by-step explanation:

In order to find the answer, first you have to find the amount of dominoes the teacher purchased by multiplying the number of packs for the number of dominoes in each pack:

7*55=385

Now, since the stament indicates that the teacher already had 118 dominoes to use in the game, you have to add this amount to the number of dominoes that were purchased:

385+118=503

According to this, the answer is that the teacher had 503 dominoes for the game.

8 0

3 years ago

Jeremy analyses one of his parachute jumps. He draws a graph showing his velocity up to the opening of his parachute. a) Estimat

jeyben [28]

Answer:

Jeremy's acceleration is about $1\,\frac{m}{s^2}$ at t =10 s

His average speed is about 44.5 m/s in this section of his jump approximating with points on the curve (under-estimate)

His average speed is about 46 m/s if using the tangent line (over estimate)

Step-by-step explanation:

Jeremy's acceleration can be estimated by the curve's derivative at that point. That is the slope of the tangent line to the velocity curve at x = 10 sec. Please see attached image where the tangent line is drawn in orange, and the points to use to calculate its slope are drawn in green.

These points are : (6, 42) and (14,50) which using the slope formula give:

$slope=\frac{y_2-y_1}{x_2-x_1}= \frac{50-42}{14-6}=\frac{8}{8} \frac{m}{s^2} = 1\,\frac{m}{s^2}$

So his acceleration at that point is about $1\,\frac{m}{s^2}$

Now, using about the same interval of x-values (from 6 to 14), the corresponding speeds are approximately: 40 (for time 6 seconds) and 49 (for time 14 seconds (look for the red dots on the attached image). Since the average velocity is given by the integral of the function between those points divided by the length of the interval where it is calculated:

$v_{average}=\frac{area}{interval\,\,length}$

and we don't have the actual velocity function to estimate the integral, we can approximate this area by that of a trapezoid that connects the red dots with the bottom of the horizontal axis (see red trapezoid in the image). Clearly from the image, this approximation would give us an under-estimate of the actual average speed.

The area of this trapezoid is: approximately:

$Trapezoid\,\, area=(49+40)\,8/2=356$

Then the average velocity estimated from it is:

$v_{average}=\frac{356}{8} \frac{m}{s} =44.5\,\frac{m}{s}$

If the area is approximated instead with the trapezoid form by the green points we used to calculate the acceleration (this would give us an over-estimate):

$Trapezoid\,\, area=(50+42)\,8/2=368$

Then the average velocity estimated from it is:

$v_{average}=\frac{368}{8} \frac{m}{s} =46\,\frac{m}{s}$

while his actual instantaneous velocity seems to be about 46 m/s from the graph

4 0

3 years ago