All categories

2 years ago

Using the slope formula find the slope of the line that passes through these points (2,5) and (8,-3)

Mathematics

2 answers:

allochka39001 [22]2 years ago

7 0

Answer:

-4/3

Step-by-step explanation:

You can use rise/run. The rise (y values) goes from 5 to -3. You subtract 8 for this. The run (x values) goes from 2 to 8. You add 6 for this. the proportion is -8/6, which can be simplified to -4/3.

Therefore, the slope is -4/3!

Hope I helped!!!!!!

zhannawk [14.2K]2 years ago

6 0

Answer:

-4/3

Step-by-step explanation:

The formula for slope is [ y2-y1/x2-x1 ].

-3-5/8-2

-8/6

-4/3

Best of Luck!

You might be interested in

$1000 is invested at 2.9% annual interest for 6 months.<br> a. Simple Interest b. Compound Interest

kirza4 [7]

Answer:

a. Simple interest = $174

b. Compund interest = $1187.1

Step-by-step explanation:

Given the following data;

Principal, P = $1000

Interest, R = 2.9%

Time, T = 6 months

a. To find the simple interest;

S.I = (PRT)/100

S.I = (1000*2.9*6)/100

S.I = 17400/100

S.I = $174

b. To find the compound interest;

$A = P(1 + \frac{r}{100})^{t}$

Substituting into the formula, we have;

$A = 1000(1 + \frac{2.9}{100})^{6}$

$A = 1000(1 + 0.029)^{6}$

$A = 1000(1.029)^{6}$

$A = 1000(1.1871)$

A = $1187.1

8 0

2 years ago

Suppose the line of best fit is being found for some data points that have an r-value of 0.657. If the standard deviation of the

oksian1 [2.3K]

We are given a line with the following data:

r-value = 0.657 (r)
standard deviation of x-coordinates = 2.445 (Sx)
standard deviation of y-coordinates = 9.902 (Sy)

We are asked to find the slope of the line up to 3 decimal places.

To find the slope of the line, based on the data that we have, we can use this formula:

slope, b = r * (Sy / Sx)

substitute the values to the formula:

b = 0.657 * ( 9.902 / 2.445 )

Solve for the b.

Therefore, the slope of the line is

b = 2.66078, round off to three decimal places:

b = 2.661 is the slope of the line.

5 0

3 years ago

Read 2 more answers

Find the six trig function values of the angle 240*Show all work, do not use calculator

-BARSIC- [3]

Solution:

Given:

$240^0$

To get sin 240 degrees:

240 degrees falls in the third quadrant.

In the third quadrant, only tangent is positive. Hence, sin 240 will be negative.

$sin240^0=sin(180+60)$

Using the trigonometric identity;

$sin(x+y)=sinx\text{ }cosy+cosx\text{ }siny$

Hence,

$\begin{gathered} sin(180+60)=sin180cos60+cos180sin60 \\ sin180=0 \\ cos60=\frac{1}{2} \\ cos180=-1 \\ sin60=\frac{\sqrt{3}}{2} \\ \\ Thus, \\ sin180cos60+cos180sin60=0(\frac{1}{2})+(-1)(\frac{\sqrt{3}}{2}) \\ sin180cos60+cos180sin60=0-\frac{\sqrt{3}}{2} \\ sin180cos60+cos180sin60=-\frac{\sqrt{3}}{2} \\ \\ Hence, \\ sin240^0=-\frac{\sqrt{3}}{2} \end{gathered}$

To get cos 240 degrees:

240 degrees falls in the third quadrant.

In the third quadrant, only tangent is positive. Hence, cos 240 will be negative.

$cos240^0=cos(180+60)$

Using the trigonometric identity;

$cos(x+y)=cosx\text{ }cosy-sinx\text{ }siny$

Hence,

$\begin{gathered} cos(180+60)=cos180cos60-sin180sin60 \\ sin180=0 \\ cos60=\frac{1}{2} \\ cos180=-1 \\ sin60=\frac{\sqrt{3}}{2} \\ \\ Thus, \\ cos180cos60-sin180sin60=-1(\frac{1}{2})-0(\frac{\sqrt{3}}{2}) \\ cos180cos60-sin180sin60=-\frac{1}{2}-0 \\ cos180cos60-sin180sin60=-\frac{1}{2} \\ \\ Hence, \\ cos240^0=-\frac{1}{2} \end{gathered}$

To get tan 240 degrees:

240 degrees falls in the third quadrant.

In the third quadrant, only tangent is positive. Hence, tan 240 will be positive.

$tan240^0=tan(180+60)$

Using the trigonometric identity;

$tan(180+x)=tan\text{ }x$

Hence,

$\begin{gathered} tan(180+60)=tan60 \\ tan60=\sqrt{3} \\ \\ Hence, \\ tan240^0=\sqrt{3} \end{gathered}$

To get cosec 240 degrees:

$\begin{gathered} cosec\text{ }x=\frac{1}{sinx} \\ csc240=\frac{1}{sin240} \\ sin240=-\frac{\sqrt{3}}{2} \\ \\ Hence, \\ csc240=\frac{1}{\frac{-\sqrt{3}}{2}} \\ csc240=-\frac{2}{\sqrt{3}} \\ \\ Rationalizing\text{ the denominator;} \\ csc240=-\frac{2}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}} \\ \\ Thus, \\ csc240^0=-\frac{2\sqrt{3}}{3} \end{gathered}$

To get sec 240 degrees:

$\begin{gathered} sec\text{ }x=\frac{1}{cosx} \\ sec240=\frac{1}{cos240} \\ cos240=-\frac{1}{2} \\ \\ Hence, \\ sec240=\frac{1}{\frac{-1}{2}} \\ sec240=-2 \\ \\ Thus, \\ sec240^0=-2 \end{gathered}$

To get cot 240 degrees:

$\begin{gathered} cot\text{ }x=\frac{1}{tan\text{ }x} \\ cot240=\frac{1}{tan240} \\ tan240=\sqrt{3} \\ \\ Hence, \\ cot240=\frac{1}{\sqrt{3}} \\ \\ Rationalizing\text{ the denominator;} \\ cot240=\frac{1}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}} \\ \\ Thus, \\ cot240^0=\frac{\sqrt{3}}{3} \end{gathered}$