∫(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.
----
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>.
D.
1/5 of 65 is
65 × 1 / 5
65 / 5
= 13
65 and 13 add up to 78.
65 is the highest number out of both numbers.
♧
243.875 rounded to the nearest tenth is...... 243.9 :)
243.875 rounded to the nearest hundredth is ...... 243.88 :)
243.875 rounded to the nearest ten is ...... 240 :)
243.875 rounded to the nearest hundred is ......200 :)
thank you for the question and hopefully this helped you :)
Answer:
ΔABD ≅ ΔACD by SAS, therefore;
by CPCTC
Step-by-step explanation:
The two column proof is presented as follows;
Statement 
Reason
ABCD is a trapezoid Given
Given
Definition of a trapezoid
ABCD is an isosceles trapezoid Left and right leg are equal
∠BAD ≅ ∠CDA Base angle of an isosceles trapezoid are congruent
Reflexive property
ΔABD ≅ ΔACD By SAS rule of congruency
CPCTC
CPCTC; Congruent Parts of Congruent Triangles are Congruent
SAS; Side Angle Side rule of congruency
Answer:
you have to provide a picture or something as well
Step-by-step explanation:
