Answer:
52 serves landed out
Explanation:
If you divide the number of serves in by 2 ( number that he lands in for every 6 serves) and then multiply by 4 ( number that land out for every 6 serves), you get 52 serves out.
Answer:
Step-by-step explanation:
We can use the Polynomial Remainder Theorem. It states that if we divide a polynomial P(x) by a <em>binomial</em> in the form (x - a), then our remainder will be P(a).
We are dividing:
So, a polynomial by a binomial factor.
Our factor is (x + k) or (x - (-k)). Using the form (x - a), our a = -k.
We want our remainder to be 3. So, P(a)=P(-k)=3.
Therefore:
Simplify:
Solve for <em>k</em>. Subtract 3 from both sides:
Factor:
Zero Product Property:
Solve:
So, either of the two expressions:
Will yield 3 as the remainder.
Answer: https://mathsolver.microsoft.com/en/solve-problem/2x+4y-8=0
Step-by-step explanation: you should find your answer there
Answer: He has 7 at the moment, technically, if he still has the bag, he still has all 36
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>.
