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3 years ago

What points satisfy 2x-6y=12 (i need 3)

Mathematics

1 answer:

kari74 [83]3 years ago

8 0

Answer:

(0,-2),(3,-1),(6,0)

Step-by-step explanation:

Plug in a number into x or y and solve the equation for the other variable. Or just use Desmos to graph it real quick and see the points.

You might be interested in

Find the radius of a circle that has an area of 6.76 cm². Use it for pi.

Inessa05 [86]

Answer:

<h2>radius = 1.47 cm</h2>

Step-by-step explanation:

Area of a circle = πr²

where

r is the radius

From the question

Area = 6.76 cm²

To find the radius substitute the value of the area into the above formula and solve for the radius

That's

$6.76 = \pi \: {r}^{2}$

Divide both sides by π

We have

${r}^{2} = \frac{6.76}{\pi} \\ r = \sqrt{ \frac{6.76}{\pi} }$

r = 1.46689291

We have the final answer as

radius = 1.47 cm

Hope this helps you

8 0

3 years ago

Consider a particle moving along the x-axis where x(t) is the position of the particle at time t, x' (t) is its velocity, and x'

vodka [1.7K]

Answer:

a) $v(t) =x'(t) = \frac{dx}{dt} = 3t^2 -12t +9$

$a(t) = x''(t) = v'(t) =6t-12$

b) $0$

c) $a(t) = x''(t) = v'(t) =6t-12$

When the acceleration is 0 we have:

$6t-12=0, t =2$

And if we replace t=2 in the velocity function we got:

$v(t) = 3(2)^2 -12(2) +9=-3$

Step-by-step explanation:

For this case we have defined the following function for the position of the particle:

$x(t) = t^3 -6t^2 +9t -5 , 0\leq t\leq 10$

Part a

From definition we know that the velocity is the first derivate of the position respect to time and the accelerations is the second derivate of the position respect the time so we have this:

$v(t) =x'(t) = \frac{dx}{dt} = 3t^2 -12t +9$

$a(t) = x''(t) = v'(t) =6t-12$

Part b

For this case we need to analyze the velocity function and where is increasing. The velocity function is given by:

$v(t) = 3t^2 -12t +9$

We can factorize this function as $v(t)= 3 (t^2- 4t +3)=3(t-3)(t-1)$

So from this we can see that we have two values where the function is equal to 0, t=3 and t=1, since our original interval is $0\leq t\leq 10$ we need to analyze the following intervals: