I REALLY NEED HELP PLEASE :)

Compare [3] and [ – 8|

mariarad [96]

Answer:

3 > -8

Step-by-step explanation:

-8 -7 -6 -5 -4 -3 -2 -1 <u>0</u> 1 2 3 ....

-8< 3

7 0

3 years ago

25 POINTS!! (EASY)MORE POINTS!!!!!!! BRAINLIEST PLSSSSSSSSSSSSSSSSSSSHELP. For Brainliest Please help)

Lostsunrise [7]

Answer:

A) 50

B) 20

C) 20

Step-by-step explanation:

To solve the first question, we can multiply both sides by 10.

Thus, we get:

6x = 300.

Then, divide by 6, to get x:

x = 50.

You can do the same thing to all of these questions.

I've already checked the answers, but if you don't trust them, just pluck the values in.

For example, let's check the answer for question A.

0.6*50 is supposed to equal 30.

Does it equal? Yes, so we know that the answer is correct.

3 0

3 years ago

Which of the following options best describes the function graphed below? A graph shows a curve that starts at the origin and go

Tamiku [17]

It's an increasing nonlinear. Exponential functions are graphed in basically any shape besides a line. This is an exponential function, therefore it is classified as nonlinear. By looking at the slope, from left to right, we can see that it is increasing, thus providing an answer that this is an increasing nonlinear function.

3 0

3 years ago

Read 2 more answers

A rock thrown vertically upward from the surface of the moon at a velocity of 36m/sec reaches a height of s = 36t - 0.8 t^2 met

Verdich [7]

Answer:

a. The rock's velocity is $v(t)=36-1.6t \:{(m/s)}$ and the acceleration is $a(t)=-1.6 \:{(m/s^2)}$

b. It takes 22.5 seconds to reach the highest point.

c. The rock goes up to 405 m.

d. It reach half its maximum height when time is 6.59 s or 38.41 s.

e. The rock is aloft for 45 seconds.

Step-by-step explanation:

Velocity is defined as the rate of change of position or the rate of displacement. $v(t)=\frac{ds}{dt}$
Acceleration is defined as the rate of change of velocity. $a(t)=\frac{dv}{dt}$

a.

The rock's velocity is the derivative of the height function $s(t) = 36t - 0.8 t^2$

$v(t)=\frac{d}{dt}(36t - 0.8 t^2) \\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\v(t)=\frac{d}{dt}\left(36t\right)-\frac{d}{dt}\left(0.8t^2\right)\\\\v(t)=36-1.6t$

The rock's acceleration is the derivative of the velocity function $v(t)=36-1.6t$

$a(t)=\frac{d}{dt}(36-1.6t)\\\\a(t)=-1.6$

b. The rock will reach its highest point when the velocity becomes zero.

$v(t)=36-1.6t=0\\36\cdot \:10-1.6t\cdot \:10=0\cdot \:10\\360-16t=0\\360-16t-360=0-360\\-16t=-360\\t=\frac{45}{2}=22.5$

It takes 22.5 seconds to reach the highest point.

c. The rock reach its highest point when t = 22.5 s

Thus

$s(22.5) = 36(22.5) - 0.8 (22.5)^2\\s(22.5) =405$

So the rock goes up to 405 m.

d. The maximum height is 405 m. So the half of its maximum height = $\frac{405}{2} =202.5 \:m$

To find the time it reach half its maximum height, we need to solve

$36t - 0.8 t^2=202.5\\36t\cdot \:10-0.8t^2\cdot \:10=202.5\cdot \:10\\360t-8t^2=2025\\360t-8t^2-2025=2025-2025\\-8t^2+360t-2025=0$

For a quadratic equation of the form $ax^2+bx+c=0$ the solutions are

$x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\mathrm{For\:}\quad a=-8,\:b=360,\:c=-2025:\\\\t=\frac{-360+\sqrt{360^2-4\left(-8\right)\left(-2025\right)}}{2\left(-8\right)}=\frac{45\left(2-\sqrt{2}\right)}{4}\approx 6.59\\\\t=\frac{-360-\sqrt{360^2-4\left(-8\right)\left(-2025\right)}}{2\left(-8\right)}=\frac{45\left(2+\sqrt{2}\right)}{4}\approx 38.41$

It reach half its maximum height when time is 6.59 s or 38.41 s.

e. It is aloft until s(t) = 0 again

$36t - 0.8 t^2=0\\\\\mathrm{Factor\:}36t-0.8t^2\rightarrow -t\left(0.8t-36\right)\\\\\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}\\\\t=0,\:t=45$

The rock is aloft for 45 seconds.

5 0

3 years ago

Please help me I DO NOT UNDERSTAND

stepladder [879]

Are they asking for the surface area?

8 0

2 years ago

I REALLY NEED HELP PLEASE :)​

I REALLY NEED HELP PLEASE :)