Answer:
y + 4 = ⅕(x + 7)
Step-by-step explanation:
The "point-slope" form of the equation of a straight line is:
y − y₁ = m(x − x₁)
where m is the slope and (x₁, y₁) is a point on the line.
If m = ⅕ and (x₁, y₁) = (-7, - 4), the point-slope equation is
y + 4 = ⅕(x + 7)
The Figure below shows the graph of your line and a point at (-7, -4).
Given the equation 4(3b + 2)² = 64,
dividing both sides of the equation by 4, we have
(3b + 2)² = 16 and getting the square root of both sides,
(3b + 2) = 4 and (3b + 2) = -4
We can solve for b for each equation and have
3b = 2 | 3b = -6
b = 2/3 | b = -2
Therefore, the values of b are 2/3 and -2 and from the choices, the answer is <span>A: b = 2/3 and b = -2.</span>
∫(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>.
