The range of the given relation is D. R = {-1, 3, 5, 8}.
Step-by-step explanation:
Step 1:
The range of a relation is the second set of values while the domain constitutes the first set of values.
There are 4 given relations with two sets of values so there would be 4 domain values and 4 range values.
Step 2:
The range of (1, -1) = -1,
The range of (2, 3) = 3,
The range of (3, 5) = 5,
The range of (4, 8) = 8.
Combining these values we get the range as {-1, 3, 5, 8} which is option D.
Y = -5x + 1
x = 1 , y = -4
x = 2 , y = -9
x = 3 , y = -14
so the answers are
(1,-4)
(2,-9)
(3,-14)
Answer:C
Step-by-step explanation:
3.1 x 10^(-4) x 8.5 x 10^(-2)
3.1 x 8.5 x 10^(-4) x 10^(-2)
26.35 x 10^(-4-2)
26.35 x 10^(-6)
2.635 x 10^1 x 10^(-6)
2.635 x 10^(1-6)
2.635 x 10^(-5)
∫(t = 2 to 3) t^3 dt
= (1/4)t^4 {for t = 2 to 3}
= 65/4.
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∫(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>.
