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3 years ago

In what grade do you learn this math equation? You don't have to solve, I'm just wondering

Mathematics

2 answers:

Sphinxa [80]3 years ago

8 0

Answer:

I think in grade 6th or 7th I star it learning those types of equations

Alexxandr [17]3 years ago

7 0

Maybe 8th or 7th grade I’m not sure

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Find the length of side x in simplest radical from with a rational denominator.

kozerog [31]

Answer:

Step-by-step explanation:

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3 years ago

Ron and Kathy are ticket sellers for their classplay. Ron sells student tickets for $2.00 each, and Kathy sells adult tickets fo

NARA [144]

$2 \div 364 =$
$182$

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3 years ago

Omar’s classroom has 2 closets. Each closet has 3 shelves. There are 5 backpacks on each shelf. How many backs are there in all

almond37 [142]

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4 years ago

The rate of change of the downward velocity of a falling object is the acceleration of gravity (10 meters/sec 2) minus the accel

Alexus [3.1K]

Answer:

See below

Step-by-step explanation:

Write the initial value problem and the solution for the downward velocity for an object that is dropped (not thrown) from a great height.

if v(t) is the speed at time t after being dropped, v'(t) is the acceleration at time t, so the the initial value problem for the downward velocity is

v'(t) = 10 - 0.1v(t)

v(0) = 0 (since the object is dropped)

<em> The equation v'(t)+0.1v(t)=10 is an ordinary first order differential equation with an integrating factor </em>

$\bf e^{\int {0.1dt}}=e^{0.1t}$

so its general solution is

$\bf v(t)=Ce^{-0.1t}+100$

To find C, we use the initial value v(0)=0, so C=-100

and the solution of the initial value problem is

$\bf \boxed{v(t)=-100e^{-0.1t}+100}$

what is the terminal velocity?

The terminal velocity is

$\bf \lim_{t \to\infty}(-100e^{-0.1t}+100)=100\;mt/sec$

How long before the object reaches 90% of terminal velocity?

90% of terminal velocity = 90 m/sec

we look for a t such that

$\bf -100e^{-0.1t}+100=90\rightarrow -100e^{-0.1t}=-10\rightarrow e^{-0.1t}=0.1\\-0.1t=ln(0.1)\rightarrow t=\frac{ln(0.1)}{-0.1}=23.026\;sec$

How far has it fallen by that time?

The distance traveled after t seconds is given by

$\bf \int_{0}^{t}v(t)dt$

So, the distance traveled after 23.026 seconds is

$\bf \int_{0}^{23.026}(-100e^{-0.1t}+100)dt=-100\int_{0}^{23.026}e^{-0.1t}dt+100\int_{0}^{23.026}dt=\\-100(-e^{-0.1*23.026}/0.1+1/0.1)+100*23.026=1,402.6\;mt$

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3 years ago

A slow jogger runs a mile in 13 mins calculate the speed in in/s

IgorC [24]

Hello!!

Well, there are 60 seconds in a minute, so all we have to do is multiply by 60.

13 * 60 = 780

The jogger runs a mile in 780 seconds.

Hope this helps!!

7 0

3 years ago