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3 years ago

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SCALCET8 3.11.501.XP. Find the numerical value of each expression. (Round your answers to five decimal places.) (a) sinh(ln(4))

(b) sinh(4)

1 answer:

raketka [301]3 years ago

6 0

sinh(ln(4)) = (exp(ln(4)) - exp(-ln(4)))/2 = (4 - 1/4)/2 = 15/8 = 1.875

sinh(4) = (exp(4) - exp(-4))/2 ≈ 27.28992

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anzhelika [568]

(f o g)(-3) = (f(g(-3))
Because g is on the inside, we carry out g first.

g(x) = x^2 - 3
Substitute -3 in for x.
g(-3) = (-3)^2 - 3 = 9 - 3 = 6
g(-3) = 6

Next, carry out f on the result of g(-3)
f(6) = 2(6) - 1
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Does the point (-2, -5) satisfy the equation y = x − -3?

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No because (-2, -5) is negative and the y coordinate needs to be positive.

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CaHeK987 [17]

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c. If a plane contains a line, it contains the points on the line.

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The only statement relating a point on a line to the plane containing the line is the one shown above.

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2 years ago

5. A tank initially contains 20 lb of salt dissolved in 200-gallon of water. Starting at time t = 0, a solution containing 0.5 l

sladkih [1.3K]

Answer:

(a) Amount of salt as a function of time

$A(t)=100-80e^{-0.02t}$

(b) The time at which the amount of salt in the tank reaches 50 lb is 23.5 minutes.

(c) The amount of salt when t approaches to +inf is 100 lb.

Step-by-step explanation:

The rate of change of the amount of salt can be written as

$\frac{dA}{dt} =rate\,in\,-\,rate\,out\\\\\frac{dA}{dt}=C_i*q_i-C_o*q_o=C_i*q_i-\frac{A(t)}{V}*q_o\\\\\frac{dA}{dt}=0.5*4-\frac{A(t)}{200}*4\\\\\frac{dA}{dt} =\frac{100-A(t)}{50}=-(\frac{A(t)-100}{50})$

Then we can rearrange and integrate

$\frac{dA}{dt}= -(\frac{A(t)-100}{50})\\\\\int \frac{dA}{A-100}=-\frac{1}{50} \int dt\\\\ln(A-100)=-\frac{t}{50}+C_0\\\\ A-100=Ce^{-0.02*t}\\\\\\A(0)=20 \rightarrow 20-100=Ce^0=C\\C=-80\\\\A=100-80e^{-0.02t}$

Then we have the model of A(t) like

$A(t)=100-80e^{-0.02t}$

(b) The time at which the amount of salt reaches 50 lb is

$A(t)=100-80e^{-0.02t}=50\\\\e^{-0.02t}=(50-100)/(-80)=0.625\\\\-0.02*t=ln(0.625)=-0.47\\\\t=(-0.47)/(-0.02)=23.5$

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3 years ago

Determine the intercepts of the line. -5x+9y=-18−5x+9y=−18minus, 5, x, plus, 9, y, equals, minus, 18 xxx-intercept: \Big((left p

natulia [17]

Answer:

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Step-by-step explanation:

To find the y-intercept, set x=0:

-5·0 +9y = -18

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To find the x-intercept, set y=0:

-5x +9·0 = -18

x = -18/-5 = 3.6

The intercepts are ...

x-intercept: 3.6

y-intercept: -2

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2 years ago