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4 years ago

Which is the best approximation to a solution of the equation e^x = 2x + 3?

1 answer:

3 0

The correct question is
Which is the best approximation to a solution of the equation
e^(2x) = 2e^{x) + 3?

we have that

e^(2x) = 2e^{x) + 3-----------> e^(2x)- 2e^{x) - 3=0
the term
e^(2x)- 2e^{x)----------> (e^x)²-2e^(x)*(1)+1²-1²------> (e^x-1)²-1

then
e^(2x)- 2e^{x) - 3=0--------> (e^x-1)²-1-3=0------> (e^x-1)²=4
(e^x-1)=2--------> e^x=3
x*ln(e)=ln(3)---------> x=ln(3)
ln(3)=1.10
hence
x=1.10

the answer is x=1.10

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3 years ago

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3 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.

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

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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

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

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

What is the fourth row of Pascal’s triangle?

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Answer:

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$formular \: for \: the \: digits \: in \: each \\ row \: of \: a \: pascal \: triangle \: is \: = 11 {}^{n} \\ row \: 0 = \: 11 {}^{0} = 1 \: \\ row \: 1 \: = 11 {}^{1} = 11 \\ row \: \: 2 \: = 11 {}^{2} = 121 \\ row \: 3 = \: 11 {}^{3} = 1331 \\ row \: 4 = 11 {}^{4} = 14641 \\ therefore \: the \: fourth \: row \: is \: 14641$

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2 years ago

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erica [24]

9514 1404 393

Answer:

sin(θ) = 4/5
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Step-by-step explanation:

The mnemonic SOH CAH TOA was created to help you remember the relevant trig relationships. It is intended to remind you ...

Sin = Opposite/Hypotenuse

sin(θ) = 4/5

Cos = Adjacent/Hypotenuse

cos(θ) = 3/5

Tan = Opposite/Adjacent

tan(θ) = 4/3

In a right triangle, the opposite side (PQ) is the one that is not part of the angle (∠R = θ). The adjacent side (QR) is the shorter side that makes up the acute angle. The longest side (PR) is the hypotenuse.

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