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3 years ago

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At an airpor, it costs $7 to park for up to one hour and $5 per hour for each additional hour. Let x respresent the number of ho

urs parked. Write the function that models teh cost C(x), of parking x hours, where x is am interger greater than 1.

1 answer:

IrinaVladis [17]3 years ago

3 0

Answer: C(x) = 5x + 7

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Determine which function has the greatest rate of change as x approaches infinity.

Mekhanik [1.2K]

Answer: h(x) = 3*x^2 - 7*x + 8

Step-by-step explanation:

The rate of change of a function is equal to the derivate:

remember that a derivate of the form:

k(x) = a*x^n is k'(x) = n*a*x^(n-1)

Then we have:

f(x) = 2*x - 10

f'(x) = 1*2* = 2

g(x) = 16*x - 4

g'(x) = 1*16 = 16

h(x) = 3*x^2 - 7*x + 8

h'(x) = 2*3*x - 1*7 = 6*x - 7

So the only that increases as x increases is h(x), this means that the greates rate of change as x approaches inffinity is the rate of change of h(x)

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3 years ago

Please someone answer my questions please

navik [9.2K]

Answer:

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Step-by-step explanation:

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3 years ago

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On Monday, Takira counted 75 splatters of bird droppings on her car. On Tuesday afternoon, she counted 21 more. What is the perc

Law Incorporation [45]

Answer:

28%

Step-by-step explanation:

percent change in the number of bird droppings = increase in splatters / splatters counted on Monday x 100

$\frac{21}{75}$ x 100 = 28%

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2 years ago

3^x = 27^a+b and a^2-b^2/(a-b)=5 What is x?

Colt1911 [192]

Your answer should be 3x-x+2=4 hope this helps..

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3 years ago

Find a linear second-order differential equation f(x, y, y', y'') = 0 for which y = c1x + c2x3 is a two-parameter family of solu

Alisiya [41]

Let $y=C_1x+C_2x^3=C_1y_1+C_2y_2$ . Then $y_1$ and $y_2$ are two fundamental, linearly independent solution that satisfy

$f(x,y_1,{y_1}',{y_1}'')=0$
$f(x,y_2,{y_2}',{y_2}'')=0$

Note that ${y_1}'=1$ , so that $x{y_1}'-y_1=0$ . Adding $y''$ doesn't change this, since ${y_1}''=0$ .

So if we suppose

$f(x,y,y',y'')=y''+xy'-y=0$

then substituting $y=y_2$ would give

$6x+x(3x^2)-x^3=6x+2x^3\neq0$

To make sure everything cancels out, multiply the second degree term by $-\dfrac{x^2}3$ , so that

$f(x,y,y',y'')=-\dfrac{x^2}3y''+xy'-y$

Then if $y=y_1+y_2$ , we get

$-\dfrac{x^2}3(0+6x)+x(1+3x^2)-(x+x^3)=-2x^3+x+3x^3-x-x^3=0$

as desired. So one possible ODE would be

$-\dfrac{x^2}3y''+xy'-y=0\iff x^2y''-3xy'+3y=0$

(See "Euler-Cauchy equation" for more info)

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3 years ago