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3 years ago

Which absolute value expression represents the distance between the points shown on the number line

Mathematics

2 answers:

RUDIKE [14]3 years ago

6 0

Answer: c

Step-by-step explanation:

The distance between the 2 numbers is 5 and -4 + -1 is -5, but because your looking for the absolute value which always makes a number positive the answer is 5.

ella [17]3 years ago

4 0

Answer:

Step-by-step explanation:

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I need a real-life reference to the mathematical definition to variables

finlep [7]

Answer: a variable is a quantity that may change within the context of a mathematical problem or experiment

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4 years ago

A circle in the xy-plane is centered at (3, 0) and has a radius with endpoint (1, 8/3 ). Which answer choice is an equation of t

gayaneshka [121]

Answer: (x - 3)² + y² = $\frac{100}{9}$

Step-by-step explanation:

Circle Formula: (x - h)² + (y - k)² = r² ; (h, k) is the center & r is the radius

Use the distance formula for (3, 0) and (1, $\frac{8}{3}$ )

$r = \sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$

$r = \sqrt{(3-1)^{2}+(0-\frac{8}{3})^{2}}$

$r = \sqrt{(2)^{2}+(-\frac{8}{3})^{2}}$

$r = \sqrt{4+\frac{64}{9}}$

$r = \sqrt{\frac{36}{9}+\frac{64}{9}}$

$r = \sqrt{\frac{100}{9}}$

$r = \frac{10}{3}$

****************************************************

Center (h, k) = (3, 0) Radius (r) = $\frac{10}{3}$

(x - h)² + (y - k)² = r²

(x - 3)² + (y - 0)² = ( $\frac{10}{3}$ )²

(x - 3)² + y² = $\frac{100}{9}$

6 0

4 years ago

A rocket is shot straight up into the air with an initial velocity of 500 ft per second and from a height of 20 feet above the g

iragen [17]

The height of the rocket above the ground after $t$ seconds is given by the equation : $H= -16t^2+Vt+h$ , where $V$ is the initial velocity and $h$ is the initial height.

Given that, $V= 500 ft/second$ and $h= 20 ft$

So, the equation will become: $H= -16t^2 +500t+20$

A) For finding the height of the rocket 3 seconds after the launch, we will plug $t=3$ into the above equation. So....

$H= -16(3)^2+500(3)+20\\ \\ H= -16(9)+1500+20\\ \\ H= -144+1500+20=1376$

So, the height of the rocket 3 seconds after the launch is 1376 feet.

B) When the rocket at a height of 400 feet, then we will plug $H= 400$

$400=-16t^2+500t+20\\ \\ 16t^2-500t-20+400=0\\ \\ 16t^2-500t+380=0\\ \\ 4(4t^2-125t+95)=0\\ \\ 4t^2-125t+95=0$

Using quadratic formula, we will get......

$t= \frac{-(-125)+/-\sqrt{(-125)^2-4(4)(95)}}{2(4)}\\ \\ t= \frac{125+/-\sqrt{14105}}{8}\\ \\ t= 30.4705... \\ and \\ t= 0.7794...$

So, after 0.7794...seconds and 30.4705...seconds the rocket is at a height of 400 feet above the ground.

C) The time duration that the rocket remains in the air means we need to find the time taken by the rocket to reach the ground. When it reaches the ground, then $H=0$ . So.....