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4 years ago

Solve the following equations for x, if possible. check your solutions. 6/dividex+2 = 3/4

Mathematics

1 answer:

SOVA2 [1]4 years ago

3 0

Solving $\frac{6}{x+2}=\frac{3}{4}$ we get x=6

Step-by-step explanation:

We need to solve $\frac{6}{x+2}=\frac{3}{4}$

Solving:

Cross multiply both terms:

$4*6=3(x+2)\\24=3x+6\\3x=24-6\\3x=18\\x=18/3\\x=6$

Verifying the solution:

Putting x=6 in the question given:

$\frac{6}{x+2}=\frac{3}{4}$

$\frac{6}{6+2}=\frac{3}{4}$

$\frac{6}{8}=\frac{3}{4}$

$\frac{3}{4}=\frac{3}{4}$

Solving $\frac{6}{x+2}=\frac{3}{4}$ we get x=6

Keywords: Solving equations

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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

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