Answer:
Option (D)
Step-by-step explanation:
Coordinates of the points J, E and V are,
J → (-4, -5)
E → (-4, -3)
V → (-1, -1)
This triangle is translated by the rule 
given in the question.
Coordinates of the image will follow the rule,
(x, y) → [(x + 2), (y + 4)]
following this rule coordinates of the image triangle will be,
J(-4, 5) → J'(-2, -1)
E(-4, -3) → E'(-2, 1)
V(-1, -1) → V'(1, 3)
Therefore, points given in the option (D) will be the answer.
Answer:
64.8
Step-by-step explanation:
maybe try multiplying the ounces by the number of nails
Answer:
yes but it would be an incomplete fraction
Step-by-step explanation:
Part A. You have the correct first and second derivative.
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Part B. You'll need to be more specific. What I would do is show how the quantity (-2x+1)^4 is always nonnegative. This is because x^4 = (x^2)^2 is always nonnegative. So (-2x+1)^4 >= 0. The coefficient -10a is either positive or negative depending on the value of 'a'. If a > 0, then -10a is negative. Making h ' (x) negative. So in this case, h(x) is monotonically decreasing always. On the flip side, if a < 0, then h ' (x) is monotonically increasing as h ' (x) is positive.
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Part C. What this is saying is basically "if we change 'a' and/or 'b', then the extrema will NOT change". So is that the case? Let's find out
To find the relative extrema, aka local extrema, we plug in h ' (x) = 0
h ' (x) = -10a(-2x+1)^4
0 = -10a(-2x+1)^4
so either
-10a = 0 or (-2x+1)^4 = 0
The first part is all we care about. Solving for 'a' gets us a = 0.
But there's a problem. It's clearly stated that 'a' is nonzero. So in any other case, the value of 'a' doesn't lead to altering the path in terms of finding the extrema. We'll focus on solving (-2x+1)^4 = 0 for x. Also, the parameter b is nowhere to be found in h ' (x) so that's out as well.
Step-by-step explanation:
Quadrilateral ABCD is inscribed in a circle.
Opposite angles of a Quadrilateral are Supplementary.
