All categories

2 years ago

What is the slope of a line parallel to slope 2 and Perpendicular to slope 2?

Mathematics

2 answers:

Rashid [163]2 years ago

5 0

Answer:

it would be right next to the other line

Step-by-step explanation:

cluponka [151]2 years ago

4 0

Parallel = same slope
It would still be slope 2

Perpendicular = opposite sign and reciprocal slope
It would be slope -1/2

You might be interested in

Consider the system of differential equations dxdt=−4ydydt=−4x. Convert this system to a second order differential equation in y

koban [17]

$\dfrac{\mathrm dy}{\mathrm dt}=-4x\implies x=-\dfrac14\dfrac{\mathrm dy}{\mathrm dt}\implies\dfrac{\mathrm dx}{\mathrm dt}=-\dfrac14\dfrac{\mathrm d^2y}{\mathrm dt^2}$

Substituting this into the other ODE gives

$-\dfrac14\dfrac{\mathrm d^2y}{\mathrm dt^2}=-4y\implies y''-16y=0$

Since $x(t)=-\dfrac14y'(t)$ , it follows that $x(0)=-\dfrac14y'(0)=4\implies y'(0)=-16$ . The ODE in $y$ has characteristic equation

$r^2-16=0$

with roots $r=\pm4$ , admitting the characteristic solution

$y_c=C_1e^{4t}+C_2e^{-4t}$

From the initial conditions we get

$y(0)=5\implies 5=C_1+C_2$

$y'(0)=16\implies-16=4C_1-4C_2$

$\implies C_1=\dfrac12,C_2=\dfrac92$

So we have

$\boxed{y(t)=\dfrac12e^{4t}+\dfrac92e^{-4t}}$

Take the derivative and multiply it by -1/4 to get the solution for $x(t)$ :

$-\dfrac14y'(t)=\boxed{x(t)=-\dfrac12e^{4t}+\dfrac92e^{-4t}}$

7 0

2 years ago

A car maintains a speed of 23 mi/h for 5 seconds. It then accelerates to a speed of 46 mi/h in 5 seconds. It maintains that spee

Vika [28.1K]

The answer is a because it starts at 0 then goes up to 23 and then it goes up to 46.

6 0

2 years ago

Read 2 more answers

PLEASE HELP WILL GIVE BRAINLIEST TO CORRECT ANSWER

nlexa [21]

Answer:

B. 6.000000 g

Step-by-step explanation:

The reason why its B is due to it having more 'sigs' than 'figs'

Hope this helped <3

6 0

2 years ago

Let us analyze the following setting. You are given a circle of unit circumference. You pickkpoints on the circle independently

garik1379 [7]

Answer:

We have been given a unit circle which is cut at k different points to produce k different arcs. Now we can see firstly that the sum of lengths of all k arks is equal to the circumference:

$\small \sum_{i = 1}^{k} L_i= 2\pi$

Now consider the largest arc to have length \small l . And we represent all the other arcs to be some constant times this length.

we get :

$\small \sum_{i = 1}^{k} C_i.l = 2\pi$

where C(i) is a constant coefficient obviously between 0 and 1.

$\small \sum_{i = 1}^{k} C_i= 2\pi/l$

All that I want to say by using this step is that after we choose the largest length (or any length for that matter) the other fractions appear according to the above summation constraint. [This step may even be avoided depending on how much precaution you wanna take when deriving a relation.]

So since there is no bias, and \small l may come out to be any value from [0 , 2π] with equal probability, the expected value is then defined as just the average value of all the samples.

We already know the sum so it is easy to compute the average :

$\small L_{Exp} = \frac{2\pi}{k}$

7 0

2 years ago

Explain how you use a net to find the surface area of a prism.

weeeeeb [17]

Answer:

Hello, Here is your answer.

Finding the areas of each of the rectangles and squares of the net of a rectangular prism and adding up those areas gives the surface area or total surface area of the prism. For example, if the length of one side of the cube 4 units then the area of one its face is 4 × 4 = 16 square units.

5 0

2 years ago