All categories

3 years ago

Slope of parallel and perpendicular line

Mathematics

1 answer:

Naya [18.7K]3 years ago

5 0

Answer:

2/5

Step-by-step explanation:

Whenever two lines are parallel, they surely have the same slope.

*extra info: whenever two lines are perpendicular to eachother, the product of their slopes will be equal to -1.

You might be interested in

A medical clinic is reducing the number of incoming patients by giving vaccines before flu season. During week 5 of flu season,

8_murik_8 [283]

Answer:

$f(x) = -6x +120$

Step-by-step explanation:

Let's call y the number of patients treated each week

Let's call x the week number.

If the reduction in the number of patients each week is linear then the equation that models this situation will have the following form:

$y = mx + b$

Where m is the slope of the equation and b is the intercept with the x-axis.

If we know two points on the line then we can find the values of m and b.

We know that During week 5 of flu season, the clinic saw 90 patients, then we have the point:

(5, 90)

We know that In week 10 of flu season, the clinic saw 60 patients, then we have the point:

(10, 60).

Then we can find m and b using the followings formulas:

$m=\frac{y_2-y_1}{x_2-x_1}$ and $b=y_1-mx_1$

In this case: $(x_1, y_1) = (5, 90)$ and $(x_2, y_2) = (10, 60)$

Then:

$m=\frac{60-90}{10-5}$

$m=-6$

And

$b=90-(-6)(5)$

$b=120$

Finally the function that shows the number of patients seen each week at the clinic is:

$f(x) = -6x +120$

7 0

3 years ago

Read 2 more answers

A projectile is fired with muzzle speed 250 m/s and an angle of elevation 45° from a position 30 m above ground level. Where doe

qaws [65]

The projectile's horizontal and vertical positions at time $t$ are given by

$x=\left(250\dfrac{\rm m}{\rm s}\right)\cos45^\circ\,t$

$y=30\,\mathrm m+\left(250\dfrac{\rm m}{\rm s}\right)\sin45^\circ\,t-\dfrac g2t^2$

where $g=9.8\dfrac{\rm m}{\mathrm s^2}$ . Solve $y=0$ for the time $t$ it takes for the projectile to reach the ground:

$30+\dfrac{250}{\sqrt2}t-4.9t^2=0\implies t\approx36.2458\,\mathrm s$

In this time, the projectile will have traveled horizontally a distance of

$x=\dfrac{250\frac{\rm m}{\rm s}}{\sqrt2}(36.2458\,\mathrm s)\approx6400\,\mathrm m$

The projectile's horizontal and vertical velocities are given by

$v_x=\left(250\dfrac{\rm m}{\rm s}\right)\cos45^\circ$

$v_y=\left(250\dfrac{\rm m}{\rm s}\right)\sin45^\circ-gt$

At the time the projectile hits the ground, its velocity vector has horizontal component approx. 176.77 m/s and vertical component approx. -178.43 m/s, which corresponds to a speed of about $\sqrt{176.77^2+(-178.43)^2}\dfrac{\rm m}{\rm s}\approx250\dfrac{\rm m}{\rm s}$ .