D C D C B are the correct answers in order
Number two is proportional just do Y over X y/x meaning divided all the y’s over the x’s so 25 divided by 5 30 divided by 6 35/7 40/8 they all equal 5 so it’s proportional
number three isn’t proportional when you do y/x they all come out differently so you do y2-y1 over x2-x1 16-4/4-2 16-4=12 4-2=2 then you divided 12/2 that equals 6 you do that again but with the next numbers but they don’t equal so it’s no equation no constant
Answer:
$3.10
Step-by-step explanation:
Answer:
The approximate circumference is 62.8 mm
Step-by-step explanation:
Circumference is basically the perimeter of the circle
Circumference is given by
Here the diameter is 20 mm, so the radius will be half of diameter = 10 mm
Substituting the given values, in above equation, we get -
Circumference =
mm
The approximate circumference is 62.8 mm
∫(t = 2 to 3) t^3 dt
= (1/4)t^4 {for t = 2 to 3}
= 65/4.
----
∫(t = 2 to 3) t √(t - 2) dt
= ∫(u = 0 to 1) (u + 2) √u du, letting u = t - 2
= ∫(u = 0 to 1) (u^(3/2) + 2u^(1/2)) du
= [(2/5) u^(5/2) + (4/3) u^(3/2)] {for u = 0 to 1}
= 26/15.
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For the k-entry, use integration by parts with
u = t, dv = sin(πt) dt
du = 1 dt, v = (-1/π) cos(πt).
So, ∫(t = 2 to 3) t sin(πt) dt
= (-1/π) t cos(πt) {for t = 2 to 3} - ∫(t = 2 to 3) (-1/π) cos(πt) dt
= (-1/π) (3 * -1 - 2 * 1) + [(1/π^2) sin(πt) {for t = 2 to 3}]
= 5/π + 0
= 5/π.
Therefore,
∫(t = 2 to 3) <t^3, t√(t - 2), t sin(πt)> dt = <65/4, 26/15, 5/π>.
