Hello from MrBillDoesMath!
Discussion:
The slope of the tangent to the curve y = 4x^3 is given by the first derivative.
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y' = 4 (3x^2) = 12x^2
At (-3,-108), the slope is 12x^2 = 12(-3)^2 = 12*9 = 108
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y = mx + b
From the first part, m = 108 so y = 108x + b.
Substituting y = -108 when x = -3 gives
-108 = 108(-3) + b => add 3* 108 to both sides
-108 + 108(3) = b => as 3 -1 = 2
108*2 = b = 216
y = 108x + 216
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Thank you,
MrB
5x + 25y = 60....multiply by -3
3x + 15y = 42....multiply by 5
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-15x - 75y = - 180 (result of multiplying by -3)
15x + 75y = 210 (result of multiplying by 5)
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0 = 30...incorrect.....this system has no solutions
Answer:
no answer.
Step-by-step explanation:
you showed 1 part of the question, add more please.
The expression x-4 will come out to be a negative integer. That means X <em>has to be </em>something that when 4 is taken away from it, the answer is still negative.
Let's try 5.
5-4 is 1. Is 1 negative? No, so that doesn't work.
4-4 is 0. 0 isn't negative either, so that doesn't work.
3-4 is -1. Is -1 negative? YES! Let's try it for 2 as well, just to be safe.
2-4 is -2. Still negative.
X can be anything less than 4.
