anonymous
  • anonymous
what exponential function passes thru (0,4) and (-2,64)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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mattfeury
  • mattfeury
the basic form for an exponential equation is \[y = ax^2 + b\] so I'd plug in (0,4) and we can solve for b then plug in (-2,64) to solve for a
mattfeury
  • mattfeury
\[4 = 0*0^2 + b\]\[b = 4\]\[64 = a*(-2)^2 + 4\]\[15 = a\] boom. plug those in your general equation (above) and you've got an answer!
anonymous
  • anonymous
these are the choices tho 1) f(x)=4(1\4)^x 2) f(x)=4^x 3)f(x)=4(1\4)^-x 4)f(x)=-4(1\4)^x

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anonymous
  • anonymous
thats why im confused?
mattfeury
  • mattfeury
well you can plug each set of numbers into all those functions and see which works. it looks like (1) will work for both sets.
mattfeury
  • mattfeury
ahh. i misread 'exponential' to be 'quadratic'. My answer is one example, but there are many functions that would work.
anonymous
  • anonymous
thank you so much and by any chance do you know this one: 3 sides of a rectangular garden are enclosed using 120 meters of fencing. what is the maximum area of garden?
mattfeury
  • mattfeury
hmm what level math are you in? I am tempted to use derivatives, but not if you haven't learned them yet.
anonymous
  • anonymous
algebra 2
anonymous
  • anonymous
and yeah i havent learned that yet lol
mattfeury
  • mattfeury
well I'm thinking that the maximum area would be a perfect square rather than a rectangle, don't you think? this would mean each of the three sides are equal.
anonymous
  • anonymous
The basic form is\[y=a^{bx}\]
myininaya
  • myininaya
ok we have x+x+y=120 so y=120-2x so A=xy=x(120-2x)=120x-2x^2 so if we find the vertex of this parabola we can find the max
anonymous
  • anonymous
I get \[y=2^{-3x}\]
myininaya
  • myininaya
i got y=4^(-3x/2) for that one
anonymous
  • anonymous
Actually, I don't...!
anonymous
  • anonymous
so with the choices 1) 3,600m^2 2) 1,800m^2 3)1,600m^2 4)900m^2 whats the correct answer
myininaya
  • myininaya
anyways the vertex of that parabola is (0,120) and you find that by writing the parabole as a*(x-h)^2+k where (h,k) is the vertex
myininaya
  • myininaya
oops one sec i made a mistake
anonymous
  • anonymous
\[y=4e^{-x \ln4}\]
anonymous
  • anonymous
If you let y be length of one of the sides, and x be the length of the other two sides, then y+2x=120. The area is given by A=xy, so since y=120-2x, just substitute for y in A=xy. This gives A=x(120-2x). You are trying to find x that maximizes this. Since the graph of this will be an upside down parabola with roots at x=0 and x=60, this means the maximum will occur at x=30 (halfway between the roots). So the max area is A=30*(120-2*30)=30*60=1800.
anonymous
  • anonymous
When x is 0, y is 4. When x is -2, y is 64.
myininaya
  • myininaya
the vertex of the parabola is (30,1800) since A=-2(x-30)^2+1800
anonymous
  • anonymous
Why are we talking about quadratics?
anonymous
  • anonymous
oh okay so the answe is 2
anonymous
  • anonymous
1800
anonymous
  • anonymous
\[y=ae^{bx}\]When x=0, y=4, so\[4=ae^0=a\]so a=4. When x=-2, y=64, so\[64=4e^{-2b} \rightarrow 16=e^{-2b} \rightarrow \ln 16 = -2b \rightarrow b=-\frac{1}{2}\ln 16=-\ln 4\]so b = -ln(4). hence your equation is,\[y=4e^{-(\ln 4)x }\]
myininaya
  • myininaya
you're so smart lokisan lol
anonymous
  • anonymous
haha...but wait, there's more...
anonymous
  • anonymous
\[e^{-(\ln4)x}=e^{\ln \frac{1}{4}x}=\left( e^{\ln \frac{1}{4}} \right)^x=\frac{1}{4^x}\]so\[y=4e^{-(\ln 4)x}=4.\frac{1}{4^x}=\frac{1}{4^{x-1}}=4^{1-x}\]
anonymous
  • anonymous
So your equation can be written as,\[y=4^{1-x}\]
anonymous
  • anonymous
Silence...
anonymous
  • anonymous
lol u guys are so good at this im so confused. but okay if thats my equation whats the answer, 1 2 3 or 4
anonymous
  • anonymous
3600 or 1800 or 1600 or 900
anonymous
  • anonymous
What do you actually need to find out? I thought you just had to come up with the equation.
anonymous
  • anonymous
imrickjames is here...haven't seen you in a while ;]
anonymous
  • anonymous
noo .. i wrote up there earlier the choices...its asking me what the maximum area of the garden is
anonymous
  • anonymous
Okay...let me check the multiple choices...
anonymous
  • anonymous
What garden? Isn't that another question?
anonymous
  • anonymous
Oh, I see the garden question. So the others didn't work it out?
anonymous
  • anonymous
no i just need to answer the garden question. i need to choose one of those 4 and show work.
anonymous
  • anonymous
1800
anonymous
  • anonymous
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anonymous
  • anonymous
Thanks for fanning me, Matt :)
anonymous
  • anonymous
thank u ..wat about if a car is purchased for 20,000. the value of the car decreases by 10% each year. about how many years will it take for the vallue of the car to reach 8,000
mattfeury
  • mattfeury
hey! you can't know it was me! heh :). impressed by your scan.
anonymous
  • anonymous
About 8.7 years.
anonymous
  • anonymous
Impressed by the scan...so *that's* how I increase the fan base ;)
anonymous
  • anonymous
how did u get that thoo
anonymous
  • anonymous
These types of equations with growth and decay in proportion to themselves follow the form\[p(t)=p(0)e^{kt}\]where p(0) is the initial value and k some constant. For growth, it's positive, for decay, negative. You get this equation by solving the differential equation,\[\frac{dp}{dt}=kp\]which is separable (I don't know what level you're at). Anyway, with this equation, you use your data to find out what p(0) and k are, then answer the question.
anonymous
  • anonymous
Here, p(0) is 20,000. To find k, you know that after 1 year, the car's value is 10% less, so it's value at time t=1 is 18,000. So,\[18,000=20,000e^{k}\]Divide both sides by 20,000 and take the natural log to find k. Now you have your p(0) and your k, so you can find p(t) at any time t.
anonymous
  • anonymous
okay thank uuu ! and if i have to find |3+i| that equals what ?
anonymous
  • anonymous
You want to find t when p(t) is 8000, so you set it up as,\[8000=20000e^{ \ln (\frac{9}{10})t}\]You have to solve for t. Divide both sides by 20,000 and again, take the natural log to find\[\frac{8000}{20000}=\ln (\frac{9}{10})t \rightarrow t =\frac{8000/20000}{\ln (9/10)} \approx 8.7yrs\]
anonymous
  • anonymous
It's asking for the magnitude. The magnitude of a complex number is z is,\[|z|=|x+iy|=\sqrt{x^2+y^2}\]You consider the real and imaginary part (note, the imaginary part is NOT iy, it's y, and y is real...yi is imaginary).
anonymous
  • anonymous
\[|3+i|=\sqrt{3^2+1^2}=\sqrt{10}\]
anonymous
  • anonymous
Good luck. If you have anymore questions, post them separately...I need to log out :D
anonymous
  • anonymous
thank youuuuu so muchhh

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