Answer:
a. dy/dx = -2/3
b. dy/dx = -28
Step-by-step explanation:
One way to do this is to assume that x and y are functions of something else, say "t", then differentiate with respect to that. If we write dx/dt = x' and dy/dt = y', then the required derivative is y'/x' = dy/dx.
a. x'·y^3 +x·(3y^2·y') = 0
y'/x' = -y^3/(3xy^2) = -y/(3x)
For the given point, this is ...
dy/dx = -2/3
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b. 2x·x' +x^2·y' -2x'·y^3 -2x·(3y^2·y') + 0 = 2x' + 2y'
y'(x^2 -6xy^2 -2) = x'(2 -2x +2y^3)
y'/x' = 2(1 -x +y^3)/(x^2 +6xy^2 -2)
For the given point, this is ...
dy/dx = 2(1 -0 +27)/(0 +0 -2)
dy/dx = -28
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The attached graphs show these to be plausible values for the derivatives at the given points.
Answer: y= 3/5 + 18
Step-by-step explanation: Bruh i legit learned this a hour ago at my tuition, ok so basically simply to slope intercept from ye? so it would be y=-3/5x+7
and then you would use the distance formula and ye distance formula is square root of x sub 2 - x sub 1 squared+ y sub 2 - y sub 1 squared
In this situation, you have:
The mode is 79 (that’s the most common number).
The median is 78.5 (the middle value or the average of the two middle values, since there is an even number of test scores).
The mean is less than both of those, as it was brought down by the few scores of 49.
The means the mean will be less than median.
Answer: C
The best way to solve this one would be simple cross multiplying. Multiply the top of one side with the bottom of the other side. In this problem you need to multiply X with 8 and multiply 3 with 5. You'll get an equation of 8X=15, then divide both sides by 8 so that you get X by itself. Your answer will them be X=1.875
