Find the slope of the line that contains the points named c (3,8),d (-2,5)

Mathematics

2 answers:

Dmitry [639]2 years ago

6 0

Answer:

3/5

Step-by-step explanation:

To find the slope, we use the formula

m = (y2-y1)/(x2-x1)

= (5-8)/(-2-3)

= -3/-5

=3/5

The slope is 3/5

faltersainse [42]2 years ago

4 0

$\bf c(\stackrel{x_1}{3}~,~\stackrel{y_1}{8})\qquad d(\stackrel{x_2}{-2}~,~\stackrel{y_2}{5}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{5-8}{-2-3}\implies \cfrac{-3}{-5}\implies \cfrac{3}{5}$

You might be interested in

Find the slope and y-intercept

Sladkaya [172]

Answer:

y-intercept:20

slope:5

Step-by-step explanation:

8 0

2 years ago

Use the explicit formula an=a1+(n-1)•d to find the 300th term of the sequence below 57,66,75,84,93

Grace [21]

$a_{300}=57+(300-1)\cdot9\\
a_{300}=57+299\cdot9\\
a_{300}=57+2691\\
a_{300}=2748$

6 0

2 years ago

Read 2 more answers

You want to get from a point A on the straight shore of the beach to a buoy which is 54 meters out in the water from a point B o

anyanavicka [17]

Answer:

$x =\dfrac{45 \sqrt{6}}{ 2}$

Step-by-step explanation:

From the given information:

The diagrammatic interpretation of what the question is all about can be seen in the diagram attached below.

Now, let V(x) be the time needed for the runner to reach the buoy;

∴ We can say that,

$\mathtt{V(x) = \dfrac{70-x}{7}+\dfrac{\sqrt{54^2+x^2}}{5}}$

In order to estimate the point along the shore, x meters from B, the runner should stop running and start swimming if he want to reach the buoy in the least time possible, then we need to differentiate the function of V(x) and relate it to zero.

i.e

The differential of V(x) = V'(x) =0

= $\dfrac{d}{dx}\begin {bmatrix} \dfrac{70-x}{7} + \dfrac{\sqrt{54^2+x^2}}{5} \end {bmatrix}= 0$