The line which passes through 2,5 and 6,8

Mathematics

1 answer:

Viefleur [7K]4 years ago

7 0

Y=3/4x+7/2 will be the line

You might be interested in

Bailey studies 3 days for her final exam in history. The first day she studied 1 1/2 hours on day 2 she studied 1/2 as long as s

borishaifa [10]

1st day: 1 1/2 hours
2nd day: half of first day = 1/2 * 1 1/2 hours
3rd day: 2 1/2 times as long as first day: 2 1/2 * 1 1/2 hours

1 1/2 + 1/2 * 1 1/2 + 2 1/2 * 1 1/2 =

= 3/2 + 1/2 * 3/2 + 5/2 * 3/2

= 3/2 + 3/4 + 15/4

= 6/4 + 3/4 + 15/4

= 24/4

= 6

Answer: She studied for 6 hours over the 3-day period.

5 0

4 years ago

Read 2 more answers

An object falls from a position of rest and

kumpel [21]

Answer:

About 4.1323 meters.

Step-by-step explanation:

We can use the following kinematic equations:

$\displaystyle v_f = v_0 + at \text{ and } \Delta d = v_0 t + \frac{1}{2} at^2$

We want to determine the distance the object has dropped after falling from rest and reaching an instantenous speed of 9 m/s.

We are given that the acceleration due to gravity is 9.8 m/s².

Using this fact and the first equation, find the time for which it took the object to reach 9 m/s. Note that the initial velocity is 0 m/s since the object started from rest.

$\displaystyle \begin{aligned} v_f = v_0 + at \\ \\ (9\text{ m/s}) & = (0\text{ m/s}) + (9.8\text{ m/s$^2$})t \\ \\ 9\text{ m/s} & = (9.8 \text{ m/s$^2$})t \\ \\ t &\approx0.9184 \text{ s} \end{aligned}$

To find how far the object dropped, we can use the second equation:

$\displaystyle\begin{aligned} \Delta d & = v_0 t + \frac{1}{2} at^2 \\ \\ & = (0\text{ m/s})(0.9184 \text{ s}) + \frac{1}{2}(9.8 \text{ m/s$^2$})(0.9184 \text{ s})^2 \\ \\ & = 4.1323 \text{ m} \end{aligned}$