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3 years ago

13

What is the slope of the line through the points (–3, 5) and (4, 5)? undefined 0 2 –2

2 answers:

Kazeer [188]3 years ago

7 0

Finding the slope using two points:

The formula for slope is

$\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

In this case you have the points (-3, 5) and (4, 5):

$y_{2} =5\\y_{1} =5\\x_{2} =4\\x_{1} =-3$

^^^Plug these numbers into the formula for slope...

$\frac{5-5}{4 - (-3)}$

$\frac{0}{7}$

0

^^^This is your slope

Hope this helped!

~Just a girl in love with Shawn Mendes

lubasha [3.4K]3 years ago

7 0

Answer:

The slope is 0.

Step-by-step explanation:

To solve this problem, first you have to use the slope formula of $\displaystyle \frac{y_2-y_1}{x_2-x_1}=\frac{rise}{run}$ .

$y_2=5\\\\y_1=5\\\\x_2=4\\\\x_1=(-3)\\$

Solve.

$\displaystyle \frac{5-5}{4-(-3)}=\frac{0}{7}=0$

In conclusion, the correct answer is 0.

<h2>Hope this helps!</h2><h2 /><h2>Have a wonderful blessing day! :)</h2><h2 /><h2>Good luck! :)</h2>

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The position at time t is $s(t) = 8t^2 - 14t -8$

Step-by-step explanation:

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Generally acceleration is mathematically represented as

$a(t) = \frac{d v}{dt }$

=> $\frac{dv}{dt} = 16$

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integrating both sides we have

$\int\limits dv = \int\limits16 dt$

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=> $c = -14$

So

$v(t) = 16 t -14$

Generally the position of the body is mathematically represented as

$s(t) = \int\limits v(t)dt$

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$s(t) = \int\limits 16t - 14 dt$

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$s(t) = 16 \frac{t^2}{2} - 14t + C$

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