Given:
Funnel is in the shape of a cone.
Radius = 4 inches
Height = 10 inches
To find:
a) The volume of the funnel
b) How many quarts of oil are in the funnel.
Solution:
Volume of cone:
in³
The volume of the funnel is 167.47 in³.
1 qt ≈ 58 in³
To convert cubic inch to quarts divide by 58.
= 2.9 quarts
Therefore 2.9 quarts of oil are in the funnel.
Answer:
C
Step-by-step explanation:
Given Coordinates:
(-4, -2), (2, -2), (4, 2), (-2, 2)
The shape that uses the coordinates (-4, -2), (2, -2), (4, 2), (-2, 2) would make a parallelogram.
See graph for answer:
So correct choice is C.
<span>an</span><span> = </span>a<span>n–1</span><span> + 21 the answer is c
</span>
Answer:
Step-by-step explanation:
Given problems are absolute value problems, So we need to plug the values of given parameters and get the final result.
We have given here,
a =-2 , b = 3 , c = -4 and d = -6
Now we know that An absolute function always gives a positive value.
Let's apply this strategy in the given problems.
1. ║a+b║
Plug a= -2 and b = 3
We get, ║-2+3║=║1║= 1
2. 5║c+b║
Plug c= -4 and b=3
i.e. 5║-4 + 3║= 5║-1║=5×1 = 5
3. a+b║c║
Plug values a= -2 , b=3 and c=-4
i.e -2 +3║-4║ = -2 + 3×4 = -2 + 12 = 10
4. ║a+c║÷(-d)
i.e ║-2 + (-4)║÷(-6) = ║-6║÷(-6) = 6÷(-6) = -1
5. 3║a+d║+b
i.e 3║-2+(-6)║+3 = 3║-8║+3 = 3×8 +3 = 27
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.
