anonymous
  • anonymous
The number of birds (b) in a park increases when the number of dogs in the park (d) decreases. Write the correct equation for this scenario and solve for the number of birds when there are 10 dogs. Dogs Birds 40 10 50 8 b = 4d; b = 40 b = 4d; b = 2.5 b = 400 over d; b= 4000 b = 400 over d b = 40
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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calculusxy
  • calculusxy
I think that it is b=4d ; b=40 Since you multiply 4*10=40 and the number of birds increases as the number of dogs decreases.
jdoe0001
  • jdoe0001
btw, dogs don't eat birds :P
jdoe0001
  • jdoe0001
well, I guess they could, wild dogs would, but domestic ones do not

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More answers

anonymous
  • anonymous
If they were starving, they would.
jdoe0001
  • jdoe0001
so, there's an assumption people go to the park with starving dogs, as opposed to well-fed dogs but just for an outting
jdoe0001
  • jdoe0001
so, one amount gets affected by the change in the other usually called a DIRECT VARIATION, that is to say that one item VARIES DIRECTLY as the other changes something like say y = 2x x = 25 y = 2(25) = 50 x = 100 y = 2(100) = 200 and so on ... so one can say that b = (SOME NUMBER) d what's that number? we dunno but we know that ------------------------ Dogs Birds 40 10 50 8 ----------------------- so when b = 40, d = 10 so let's plug that in b = (SOME NUMBER) d 40 = (SOME NUMBER) 10 now just solve for "SOME NUMBER"
jdoe0001
  • jdoe0001
hmmm
jdoe0001
  • jdoe0001
anyhow, I notice my 1st figure didn't match the 2nd.... so it an inverse one, so lemme rewrite what I said
jdoe0001
  • jdoe0001
in your case is pretty much the same much the same thing, just inverse so called INVERSE VARIATION that is something VARIES INVERSELY as changes occur to another it'd be written as \(\bf b = \cfrac{\textit{some number}}{d}\) same as before, we dunno what the number is but we know that ------------------------ Dogs Birds 40 10 50 8 ----------------------- when d = 40, b = 10 so \(\bf b = \cfrac{\textit{some number}}{d} \implies 40 = \cfrac{\textit{some number}}{10}\) solve for "some number"
ybarrap
  • ybarrap
$$ m={\triangle\text{Birds}\over\triangle\text{Dogs}}={8-10\over50-40}=\dfrac{-1}{5}\\ B(D)={-1\over5}D+B_i\text{, where }B_i\text{ is the number of Birds where there are no Dogs. }\\ B(40)=10={-1\over5}(40)+B_i\implies B_i=18\\ \text{So, }B(D)={-1\over5}D+18\text{.}\\ B(10)={-1\over5}(10)+18=40.\\ \text{There would be 40 Birds when there are 10 Dogs.} $$
anonymous
  • anonymous
Thanks for the help guys! And @jdoe0001, I wasn't insinuating that a healthy, well fed dog would eat a bird. I meant a homeless dog with no food.
ILoveComputers1
  • ILoveComputers1
@jdoe0001 So would the last one be the answer?
jdoe0001
  • jdoe0001
hmm what do you mean? for the slope exercise?
ILoveComputers1
  • ILoveComputers1
Never mind I figured it out. Thank you though :) @jdoe0001
kaylaprincess
  • kaylaprincess
did you ever think that the dogs would bark at the birds, lmao you guys go straight to killings lol
anonymous
  • anonymous
ikr

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