In a quadratic equation
q(x) = ax^2 + bx + c
The discriminant is = b^2 - 4ac
We have that discriminant = 3
If
b^2 - 4ac > 0, then the roots are real.
If
b^2 - 4ac < 0 then the roots are imaginary
<span>In
this problem b^2 - 4ac > 0 3 > 0 </span>
then
the two roots must be real
Answer:
Step-by-step explanation:
so when you post your picture you wanna click the thing that looks like a paper clip and attach a file or picture. or your can copy your photo and paste it in your question if you dont know how to take a picture just screenshot it *depends what kind of computer your using*
A. The number of fabric-pattern-color combinations is 4 * 13 * 9 = 468
B. P(1st choice) = no of novels / total books = 3/6 = 1/2
P(2nd choice) = no of remaining novels/ total remaining books = 2/5
P(both novels) = 1/2 * 2/5 = 1/5 (without replacement assumed)
C. P(1st choice) = no of biographies / total books = 2/6 = 1/3
P(2nd choice) = no of remaining biographies/ total remaining books = 1/5
P(both biographies) = 1/3 * 1/5 = 1/15 (without replacement assumed)
D. P(1st choice) = no of history books / total books = 1/6
P(2nd choice) = no of novels/ total remaining books = 3/5
P(a history, then a novel) = 1/6 * 3/5 = 1/10 (without replacement assumed)
∫(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>.
18+11v=w-13t
18+11-w=-13t
13t=w-11v-18
t=1/13w - 11/13v - 1 5/13
Hope this helps :)
