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Lyrx
7 days ago
10

A sum lent out at simple interest becomes rs4480 in 3 years and rs 4800 in 5years.find the rate of interest​

Mathematics
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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
2 months ago
Read 2 more answers
Sal's Sandwich Shop sells wraps and sandwiches as part of its lunch specials. The profit on every sandwich is $2 and the profit
zzz [12365]
This question is quite lengthy, so I will offer a brief overview. To find your x-value, set y to 0 in the given equation.

2x + 3y = 1470

2x + 3(0) = 1470

2x = 1470

x = 735

Thus, your farthest point on the x-axis is (735,0).

Now, repeat for y.

2x + 3y = 1470.

2(0) + 3y = 1470

3y= 1470

y= 490

Your highest point thus is (0,490).

After plotting both points, connect them with a straight line to illustrate your graph.

Check
8 0
1 month ago
Below is an attempt to derive the derivative of sec(x) using product rule, where x is in the domain of secx. In which step, if a
tester [12383]

The mistake is present in step 3. According to the product rule, we find

\dfrac{\mathrm d}{\mathrm dx}(\sec x\times\cos x)=\dfrac{\mathrm d}{\mathrm dx}(\sec x)\times\cos x+\sec x\times\dfrac{\mathrm d}{\mathrm dx}(\cos x)

=\dfrac{\mathrm d}{\mathrm dx}(\sec x)\times\cos x\boxed{+\sec x\times(-\sin x)}

=\dfrac{\mathrm d}{\mathrm dx}(\sec x)\times\cos x\boxed{-\sec x\times\sin x}

(meaning that a factor of \sin x is overlooked)

Then

\dfrac{\mathrm d}{\mathrm dx}(\sec x)\times\cos x-\sec x\times\sin x=0

\implies\dfrac{\mathrm d}{\mathrm dx}(\sec x)\times\cos x=\sec x\times\sin x

\implies\dfrac{\mathrm d}{\mathrm dx}(\sec x)=\dfrac{\sec x\times\sin x}{\cos x}

\implies\dfrac{\mathrm d}{\mathrm dx}(\sec x)=\sec x\times\tan x

6 0
1 month ago
Initially 5 grams of salt are dissolved into 10 liters of water. Brine with concentration of salt 5 grams per liter is added at
Svet_ta [12734]

The salt enters at a rate of (5 g/L)*(3 L/min) = 15 g/min.

The salt exits at a rate of (x/10 g/L)*(3 L/min) = 3x/10 g/min.

Thus, the total rate of salt flow, represented by x(t) in grams, is defined by the differential equation,

x'(t)=15-\dfrac{3x(t)}{10}

which is linear. Shift the x term to the right side, then multiply both sides by e^{3t/10}:

e^{3t/10}x'+\dfrac{3e^{t/10}}{10}x=15e^{3t/10}

\implies\left(e^{3t/10}x\right)'=15e^{3t/10}

Next, integrate both sides and solve for x:

e^{3t/10}x=50e^{3t/10}+C

\implies x(t)=50+Ce^{-3t/10}

Initially, the tank contains 5 g of salt at time t=0, so we have

5=50+C\implies C=-45

\implies\boxed{x(t)=50-45e^{-3t/10}}

The duration required for the tank to contain 20 g of salt is t, such that

20=50-45e^{-3t/10}\implies t=\dfrac{20}3\ln\dfrac32\approx2.7031\,\mathrm{min}

3 0
2 months ago
Madeline usually makes 85% of her shots in basketball. if she attempts 20, how many will she likely make?​
Leona [12618]
17Step-by-step explanation:To determine how many shots Madeline is likely to make, divide 20 by 100 and then multiply by 85.
6 0
2 months ago
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