The answer is correct but i do not know what ARLENE did wrong
Answer:
3rd option: 60 degrees
Step-by-step explanation:
We can see in the diagram that the angle on C is a supplementary angle, which means that the sum of 135 and internal angle will be equal to 180 degrees.
Let x be the internal angle,
Then
x+135 = 180
x = 180-135
x = 45 degrees
So now we know that two interior angles of the triangle.
Also we know that sum of all internal angles of triangle is 180 degrees.
Using the same postulate:
A+B+C = 180
75 + B + 45 = 180
120+B = 180
B = 180 - 120
B = 60 degrees
So,
third option is the correct answer ..
Answer:
(x+2) (x+3) (x-5)
Step-by-step explanation:
x³-19x-30 = (x+2) (x²+ax-15) ... x³=x*(1*x²) while -30= (2)*(-15)
x³ +<u> 0</u>*x² - 19x -30 = x³ + (<u>2+a</u>)x² + (2a-15)x -30
2+a = 0
a = -2
x³-19x-30 = (x+2) (x²-2x-15) = (x+2) (x+3) (x-5)
Answer:
Vota Trump
Step-by-step explanation:
∫(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>.
