What is the slope of the line that contains ( 5, -3) and 6, 7)? Write your answer as m=

nataly862011 [7]

Answer: $-\frac{8}{5}$

Remember: RISE/RUN (y/x). Lines that are increasing have a positive slope, and lines that are decreasing have a negative slope.

You can find the slope in two ways:

1. Useful if the line is graphed: count the units between 2 points on the line.

Let's use the points (-1, 4) and (4, -4).

(-1, 4) is 8 units higher than (4, -4) and 5 units to the left of (4, -4).

Because the line is decreasing, the slope is negative.

Therefore, the slope is $-\frac{8}{5}$ .

2. Useful if the line is not graphed: find the difference between the y-coordinate values divided by the difference of the x-coordinate values.

Let's use the points (-1, 4) and (4, -4).
$\frac{-4 - 4}{4 - (-1)} = \frac{-8}{5}$

Therefore, the slope is $-\frac{8}{5}$ .

4 0

4 years ago

Read 2 more answers

Suppose that you are a city planner who obtains and sample of 20 randomly selected members of a mid-sized town in order to deter

Dmitry_Shevchenko [17]

Answer:

The critical value for the 95% confidence interval is z = 1.96.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.95}{2} = 0.025$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$

So it is z with a pvalue of $1-0.025 = 0.975$ , so z = 1.96.

The critical value for the 95% confidence interval is z = 1.96.

5 0

3 years ago

Can someone explain how this is right please

Daniel [21]

Because when you add them all up that's what the answers are

4 0

3 years ago

On Mars the acceleration due to gravity is 12 ft/sec^2. (On Earth, gravity is much stronger at 32 ft/sec^2.) In the movie, John

insens350 [35]

Solution :

Given initial velocity, v= 48 ft/s

Acceleration due to gravity, g = $12\ ft/s^2$

a). Therefore the maximum height he can jump on Mars is

$H_{max}=\frac{v^2}{2g}$

$H_{max} = \frac{(48)^2}{2 \times 12}$

= 96 ft

b). Time he can stay in the air before hitting the ground is

$T=\frac{2v}{g}$

$T=\frac{2 \times 48}{12}$

= 8 seconds

c). Considering upward motion as positive direction.

v = u + at

We find the time taken to reach the maximum height by taking v = 0.

v = u + at

0 = 16 + (12) t

$t=\frac{16}{12}$

$=\frac{4}{3} \ s$

We know that, $S=ut + \frac{1}{2}at^2$

Taking t = $=\frac{4}{3} \ s$ , we get

$S=16 \times\frac{4}{3} + \frac{1}{2}\times(-12) \times \left(\frac{4}{3}\right)^2$

$S=\frac{32}{3}$ feet

Thus he can't reach to 100 ft as it is shown in the movie.

d). For any jump whose final landing position will be same of the take off level, the final velocity will be the initial velocity.

Therefore final velocity is = -16 ft/s

3 0

3 years ago

Barbara found that six of the 24 students in her class on a cell phone what is the ratio of students that only sell phone to stu

4vir4ik [10]

Answer:

6:24 or 1:4

Step-by-step explanation:

If you don't have to simplify, then the answer would be 6:24, but if you do, then the answer would be 1:4.

7 0

3 years ago