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MrRissso
13 days ago
10

Let a and b be two positive numbers. If 2a + 3b=6 then the maximum product of these a and b is:

Mathematics
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Harry owes the bank money. To repay his debt, he paid \$150$150dollar sign, 150 back to the bank each month. After 101010 months
Zina [12379]


$8,400
Explanation:
Monthly repayment = $150
Total Months of repayment = 10 months
Remaining balance after 10 months = $6900
Cumulative payment = $150 × 10 = $1500
Balance remaining = $6900
Overall debt amount:
(Remaining amount + total repaid)
$(6900 + 1500)
7 0
2 months ago
A statistics professor at a large university hypothesizes that students who take statistics in the morning typically do better t
Svet_ta [12734]

Answer:

The resulting value is 2.381

Step-by-step explanation:

Using the information presented in the question, we will calculate the evidence supporting the professor's hypothesis

Given that:

x₁ = 74,

n₁ = 36

s₁ = 8

x₂ = 68

n₂ = 36

s₂ = 10

The hypotheses can be outlined as:

The critical value is t₃₆+₃₆-₂,₀.₀₁ = t₇₀,₀.₀₁

thus,

t₃₆+₃₆-₂,₀.₀₁ = t₇₀,₀.₀ = 2.381

This implies a range of -2.381 to 2.381

Thus, we can support the professor's assertion.

5 0
2 months ago
Consider the vibrating system described by the initial value problem. (A computer algebra system is recommended.) u'' + u = 8 co
PIT_PIT [12445]

Answer:

u(t)  = -(3 + w^2 ) cos t /(1- w^2)cos t + 7 sin t + 8 cos wt /(1- w^2)

Step-by-step explanation:

The characteristic equation is k² + 1 = 0, which leads to k² = -1, resulting in k = ±i.

The roots are k = i or -i.

The general solution takes the form  u(x)=C₁cosx+C₂sinx.

Applying the method of undetermined coefficients, we have

Uc(t) = Pcos wt  + Qsin wt

Calculating the derivatives gives us Uc’(t) = -Pwsin wt  + Qwcos wt

And differentiating again yields Uc’’(t) = -Pw^2cos wt  - Qw^2sin wt

With the equation U’’ + u = 8cos wt, we substitute:

-Pw^2cos wt  - Qw^2sin wt + Pcos wt  + Qsin wt = 8cos wt.

This simplifies to (-Pw^2 + P) cos wt   + (-Qw^2 + Q) sin wt = 8cos wt.

From -Pw^2 + P = 8, we find P= 8  /(1- w^2).

From -Qw^2 + Q = 8, we can conclude Q = 0.

Thus, Uc(t) = Pcos wt  + Qsin wt = 8 cos wt /(1- w^2).

Combining gives us U(t) = uh(t ) + Uc(t)

     = C1cos t + c2 sin t + 8 cos wt /(1- w^2).

Initial conditions yield:

U(0) = C1cos(0) + c2 sin (0) + 8 cos (0) /(1- w^2)

Which leads us to C1 + 8 /(1- w^2) = 5

So C1 = 5 - 8 /(1- w^2) = -(3 + w^2 ) /(1- w^2).

Next, taking the derivative:

U’(t) = -C1 sin t + c2 cos t - 8 w sin wt /(1- w^2).

Evaluating at t = 0 gives us:

U’(0) = -C1 sin (0) + c2 cos (0) - 8 w sin (0) /(1- w^2) = 7.

Thus, c2 = 7.

  u(t)  = -(3 + w^2 ) cos t /(1- w^2)cos t + 7 sin t + 8 cos wt /(1- w^2)

   

4 0
3 months ago
Read 2 more answers
The function f(x)=-x^2-2x+15 is shown on the graph. What are the domain and range of the function?
PIT_PIT [12445]

Réponse:

Le domaine est l'ensemble des nombres réels

y ≤ 16

Explication étape par étape:

Le domaine est l'ensemble des valeurs x que le graphe couvre. Il n'y a aucune restriction sur les valeurs x. Le domaine est l'ensemble des nombres réels.

La plage est l'ensemble des valeurs y que le graphe couvre. Les valeurs y atteignent au maximum 16 au sommet de la parabole. Ainsi, la plage est y ≤ 16.

3 0
2 months ago
Read 2 more answers
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