Verify that the given point is on the curve and find the lines that are (a) tangent and (b) normal to the curve at the given point.
I. x2 + xy - y2 = 1, (2,3) (10)
II x2 + y2 = 25, (3,-4)
III. x3y2 = 9, (-1,3)
IV. y2 – 2x – 4y - 1 = 0, (-2, 1)

For each curve, plug in the given point and check if the equality holds. For example:

(I) (2, 3) does lie on since 2^2 + 2*3 - 3^2 = 4 + 6 - 9 = 1.

For part (a), compute the derivative , and evaluate it for the given point . This is the slope of the tangent line at the point. For example:

(I) The derivative is

so the slope of the tangent at (2, 3) is

and its equation is then

For part (b), recall that normal lines are perpendicular to tangent lines, so their slopes are negative reciprocals of the slopes of the tangents, . For example:

(I) The tangent has slope 7/4, so the normal has slope -4/7. Then the normal line has equation

the answer would be 150 books

step-by-step explanation:

if you divide 300 by 2 you will get 150

d. 24+-0.114

step-by-step explanation:

the question requires us to calculate the approximate 95% confidence interval for the mean number of ounces of ketchup per bottle in the sample given the following statistics;

sample mean = 24

standard deviation, σ = 0.4

sample size , n = 49

the confidence interval for a population mean is calculated using the formula;

sample mean ± z-score*(σ/sqrt(n))

the z-score associated with a 95% confidence interval, from the standard normal tables, is;

1.96

substituting these values into the formula;

24 ± 1.96*(0.4/7)

= 24 ± 0.112

from the alternatives given, choice d is closest to the above expression;