All categories

2 years ago

What is the line of best fit for these points? (0,6)(2, 10)(-2, 2)(-1,4)

Mathematics

1 answer:

Likurg_2 [28]2 years ago

7 0

Answer:

Connect the dots, just label them on the graph and draw a line almost like connecting the dots

You might be interested in

Instructions: Using the image, find the distance between the points given on the graph.

9966 [12]

Answer:

d = 2.54951

Step-by-step explanation:

For:

(X1, Y1) = (4, 2)

(X2, Y2) = (4.5, 4.5)

d = \sqrt {(4.5 - 4)^2 + (4.5 - 2)^2}

6 0

2 years ago

What cryptographic method, first proposed in the mid-1980s, makes use of sloping curves instead of large prime numbers?

bearhunter [10]

Answer:

waffaafa

Step-by-step explanation:

awfafafafafaf

4 0

3 years ago

What is (+,+) reflected over the x axis?

sleet_krkn [62]

Answer:

means the the x-axis is positive and the y-axis is positive so it is located in the first quarter of the cariesian corrdenates

5 0

2 years ago

Read 2 more answers

How to do slope intercept form again?!

earnstyle [38]

I don't know what is the correct answer

3 0

2 years ago

Yea I can do tomorrow it would take me to get the money back to you

Tems11 [23]

Given, the equation that represents the height of an object:

$y(t)=100t-16t^2$

First, we will find the velocity of the object which is the first derivative of the height using the method of the limits

$\frac{dy}{dt}=\lim_{h\to a}\frac{y(3+h)-y(3)}{(3+h)-(3)}$

We will find the value of the function y(t) when t = 3, and when t = 3+h

$\begin{gathered} y(3+h)=100(3+h)-16(3+h)^2 \\ y(3+h)=300+300h-16(9+6h+h^2) \\ y(3+h)=300+300h-144-96h-16h^2 \\ y(3+h)=156+4h-16h^2 \\ y(3)=100(3)-16(3)^2=156 \end{gathered}$

Substitute y(3+h) and y(3) into the expression of the limit

$\begin{gathered} \frac{dy}{dt}|_{t=3}=\lim_{h\to a}\frac{156+4h-16h^2-156}{3+h-3}=\lim_{h\to a}\frac{4h-16h^2}{h} \\ \\ \frac{dy}{dt}|_{t=3}=\lim_{h\to a}(4-16h) \end{gathered}$

Where a = 0

d) compute the instantaneous velocity at t = 3

$\frac{dy}{dt}|_{t=3}=100-16*2*3=4$

So, the answer will be:

$\begin{gathered} \frac{dy}{dt}|_{t=3}=\lim_{h\to a}(4-16h) \\ a=0 \\ \\ \frac{dy}{dt}|_{t=3}=4\text{ ft/sec} \end{gathered}$

5 0

1 year ago