Answer:
yesssss u are
Step-by-step explanation:
Find the horizontal distance of 230 and find the Vertical distance , which is where the black dot is located.
The black dot is on 49 inches.
Now find the vertical distance f the black line at horizontal 230: This is on 47.5.
The difference between the two is : 49 - 47.5 = 1.5
The answer would be A. 1.5
B is correct ..............
You can use substitution for this.
x+5 = 5x -11
Combine like terms
16=4x
Divide 4 on both sides to isolate the x
x=4
Plug x=4 to one of the equations
y= 4 +5
Combine like terms
y=9
So the answer is A. (4,9)
I hope this helps you out a bunch hun :)
∫(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>.