Simplify. 1/4√8n √2n^3

Cesar has read 89 pages of a book for his

slega [8]

Answer:

X=8

Step-by-step explanation:

First subtract the read pages with the unread pages which is 168.

Then let 21 pages = to one hour then 168pages(the remaining pages) = x then cross multiply and divide then your answer will be 8

4 0

3 years ago

Read 2 more answers

The scale for the drawing of a rectangular playing field is 2 inches=7 feet. Find an equation you can use to find the dimensions

lina2011 [118]

Answer:

The actual answer is 7/2

Step-by-step explanation:

5 0

3 years ago

Marcy played the piano for 45 minutes. She stopped playing at 4:15 p.m. At what time did she start playing the piano?

fiasKO [112]

Well, first off 1 hr =60 minutes. 60 min - 45 min = 15 minutes extra. what is one hour before 4:15? 3:15! Add the subtracted 15 minutes to get 3:30... Hope this helps

4 0

3 years ago

Read 2 more answers

Which graph represents the solution set of the system of inequalities?

Nimfa-mama [501]

Well first we need to change the format of the equations to slope-intercept, or y=mx+b.
So the first one (x + y < 1) will be changed to y < -x + 1.
The second one (2y ≥ x - 4) will be changed to y <span>≥ x/2 - 2.
Now we can analyze each graph.
In every single graph the first equation (y < -x + 1) is graphed correctly.
Now for the second equation, we can see that only the first and last graph correctly format to the equation.
Now for the shading:
The first equation shows us that y is less than -x +1, making the shading go under the dotted line. (to the left)
The second equation shows us that y is greater than or equal to x/2 - 2, making the shading go above the line. (also to the left)
Therefore, when we shade, the overlapping shading is correctly formatted in the first graph.
Hope this helped, comment any questions you have for me.</span>

5 0

3 years ago

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