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Question 103300: I have two problems. I worked part of one but cannot get it finished and don't know why. It is
-16t^2 + 112t = 120
-16t^2 + 112t - 120 = 0
t=(-112±√(〖112〗^2-4∙(-16)∙(-120) ))/(2∙(-16) )
t=(-112±√(12544-7680))/(-32)
t= (-112±√4864)/(-32)
And the other is Find the distance between (–4, 0) and (–5, –3)I put...
-5+(-4)=-9
-3+0=-3
-9^2+-3^2=h^2
81+9=h^2
h=sqrt90
h=sqrt9*10
h=3sqrt10
Where did I go wrong????
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Find the distance between (–4, 0) and (–5, –3)
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Formula:
distance = sqrt[(change in y)^2 + (change in x)^2]
distance = sqrt[(-3-0)^2 + (-5--4)^2]
distance = sqrt[9 + 1]
distance = sqrt(10)
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Cheers,
Stan H.
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