Answer:
I'm going to put the most shells possible on the necklaces this would be 38, 38x5=190, meaning that we have a remainder of 2.
2 shells would be on the magnet
Answer:
34.43
Step-by-step explanation:
A differential of length in terms of t will be ...
dL(t) = √(x'(t)^2 +y'(t)^2)
where ...
x'(t) = 4cos(4t)
y'(t) = 7cos(7t)
The length of c(t) will be the integral of this differential on the interval [0, 2π].
Dividing that interval into 10 equal pieces means each one has a width of (2π)/10 = π/5. The midpoint of pieces numbered 1 to 10 will be ...
(π/5)(n -1/2), so the area of the piece will be ...
sub-interval area ≈ (π/5)·dL((π/5)(n -1/2))
It is convenient to let a spreadsheet or graphing calculator do the function evaluation and summing of areas.
__
The attachment shows the curve c(t) whose length we are estimating (red), and the differential length function (blue) we are integrating. We use the function p(n) to compute the midpoint of the sub-interval. The sum of sub-interval areas is shown as 34.43.
The length of the curve is estimated to be 34.43.
So, we have the two equations $4x+$360, and $10x+$6x. x represents how much members pay each day. The total number of visits when the cost for a member and nonmember is the same means we will have to set both equations equal to one another.
<span>$4x+$360=$10x+$6x </span>
<span>Combine like terms. </span>
<span>$4x+$360=$16x </span>
<span>Subtract $4x on both sides to balance out the equation. </span>
<span>$12x=$360 </span>
<span>Divide by $12.00 on each side. </span>
<span>x=30 visits </span>
<span>To check your work, plug in what x equals or 30 in the original equation. If you come up with a true statement, then you know your answer has to be correct. </span>
<span>4x+360=16x </span>
<span>4(30)+360=16(30) </span>
<span>120+360=480 </span>
<span>480=480 </span>
<span>Because this is a true statement, you can be certain that the total number of visits when the cost of a member and a nonmember will be the same is 30 visits. Hope I helped! Brainliest too please.</span>