23 is the outerlier.......
Answer:
x = 5
Step-by-step explanation:
Using Pythagoras' identity in the right triangle.
The square on the hypotenuse is equal to the sum of the squares on the other 2 sides, that is
x² + 12² = 13², so
x² + 144 = 169 ( subtract 144 from both sides )
x² = 25 ( take the square root of both sides )
x = 
= 5
If you solve for x the answer is x<-8
483,600,000 miles
4 - hundred millions place
8 - ten millions place
3 - millions place
6 - hundred thousands place
0 - ten thousands, thousands, hundreds, tens, and ones place
∫(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>.
