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2 years ago

Can someone help me with this problem ?

Mathematics

2 answers:

Flura [38]2 years ago

6 0

Answer:

Step-by-step explanation:

You would subtract 5 from 88 to get Anjali's current age which is 83.

Alecsey [184]2 years ago

5 0

The answer is 83. 83+5=88. just subtract it.

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Would you please help me to solve this 9.1x10^-5)^-4

kap26 [50]

[( 91 * 10^-1 ) *10^ -5]^-4 = (91 * 10^-6)^-4 = 91^-4 * 10^24 = ( 10^24 )/ ( 91^4 ) =
10000000 ≈ 1458258211769

7 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. $a(t)=\frac{dv}{dt}$

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

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