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3 years ago

Which of the following ordered pairs is a solution of 3x+2y=-19

1 answer:

34kurt3 years ago

8 0

We need the ordered pairs to choose an answer, but if you want to do it yourself you plug in the x and y values (x,y) and test to see if the equation is true, and equals -19.

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A fan blade spins at a constant rate. This spinning is known as revolutions. The equation y=750x represents the number of revo

Aneli [31]

Answer:

(0,0)

(1, 750)

(2, 1500)

(3, 2250)

Step-by-step explanation:

To list the ordered pairs (x,y) we need to calculated the number of revolutions for minutes 0, 1, 2 and 3.

To calculated the number of revolutions for 0 minutes, we need to replace x by 0 in the initial equation and solve for y as:

y = 750*x

y = 750*0

y = 0

It means that our first ordered pair is (0,0), because when we replace x by 0 we get y equal to 0

At the same way, if we replace x by 1, we get:

y = 750*x

y = 750*(1)

y = 750

So, our second ordered pair is (1, 750)

Then, replacing x by 2, we get:

y = 750*x

y = 750*(2)

y = 1500

So, our third ordered pair is (2, 1500)

Finally, replacing x by 3, we get:

y = 750*x

y = 750*(3)

y = 2250

So, our last ordered pair is (3, 2250)

6 0

3 years ago

I’ll put you as brainliest

kari74 [83]

The law of an object moving with constant acceleration is

$s(t)=s_0+v_0t+\dfrac{1}{2}at^2$

Where $s$ is space, $t$ is time, $s_0$ is the initial position, $v_0$ is the initial velocity and $a$ is the acceleration.

In this case, if we choose a reference grid with the vertical axis pointing upwards, the acceleration of gravity will point downwards (and thus be negative). The initial position is zero, because the rocket is on the ground, and the initial velocity is 100 (positive because pointing upwards).

So, its law is

$h(t)=100t-\dfrac{1}{2}gt^2$

(I changed $s$ for $h$ since the rocket is moving vertically, so its position is actually its height. Also, g is the acceleration due to gravity).

The rocket hits the ground if its height is zero, so if we set $h=0$ we have

$0=100t-\dfrac{1}{2}gt^2 \iff t(100-\dfrac{1}{2}gt)=0$

Solving for t, we have either t=0, or

$100-\dfrac{1}{2}gt=0 \iff 100=\dfrac{1}{2}gt \iff 200=gt \iff t=\dfrac{200}{g}$

The solution t=0 means that at the beginning the rocket is on the ground. So, we're interested in the other solution. Considering that g is about 32.2 feet/s^2, we have

$t=\dfrac{200}{g}\approx \dfrac{200}{32.2}\approx 6.21$