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2 years ago

11

a boy who is reading 20 pages of a book per day can finish it in one month. how many day will he take it if he reads 30 pages pe

r day

1 answer:

Natalka [10]2 years ago

8 0

Answer:

20

Step-by-step explanation:

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Two ways to do a number line

lana66690 [7]

You can use a number line to add and subtract

8 0

2 years ago

These are the fees for sending a package by Media Mail:

Serga [27]

For first pound it is = $2.41
For next six, = 0.41 * 6 = $2.46

Remaining = 7.99 - (2.41 + 2.46) = 7.99 - 4.87 = 3.12

Now, additional pounds = 3.12 / 0.39 = 8
Total weight = 1 + 6 + 8 = 15

In short, Your Answer would be 15 pounds

Hope this helps!

6 0

3 years ago

In triangle JKL, the length of side JK is 12 feet, and the length of side KL is 15 feet. Which of the following could be the len

Andrei [34K]

Answer:

D

Step-by-step explanation:

every 2 sides of a triangle has to be bigger than the 3rd side

8 0

3 years ago

I need help with this please thank you very much

Hitman42 [59]

So,

We can notice that the graph of g, is translated 2 units to the left and 4 units up. We can express these changes with the following equation:

$g(x)=(x+2)^2+4$

4 0

9 months ago

Consider the differential equation:

Wewaii [24]

(a) Take the Laplace transform of both sides:

$2y''(t)+ty'(t)-2y(t)=14$

$\implies 2(s^2Y(s)-sy(0)-y'(0))-(Y(s)+sY'(s))-2Y(s)=\dfrac{14}s$

where the transform of $ty'(t)$ comes from

$L[ty'(t)]=-(L[y'(t)])'=-(sY(s)-y(0))'=-Y(s)-sY'(s)$

This yields the linear ODE,

$-sY'(s)+(2s^2-3)Y(s)=\dfrac{14}s$

Divides both sides by $-s$ :

$Y'(s)+\dfrac{3-2s^2}sY(s)=-\dfrac{14}{s^2}$

Find the integrating factor:

$\displaystyle\int\frac{3-2s^2}s\,\mathrm ds=3\ln|s|-s^2+C$

Multiply both sides of the ODE by $e^{3\ln|s|-s^2}=s^3e^{-s^2}$ :

$s^3e^{-s^2}Y'(s)+(3s^2-2s^4)e^{-s^2}Y(s)=-14se^{-s^2}$

The left side condenses into the derivative of a product:

$\left(s^3e^{-s^2}Y(s)\right)'=-14se^{-s^2}$

Integrate both sides and solve for $Y(s)$ :

$s^3e^{-s^2}Y(s)=7e^{-s^2}+C$

$Y(s)=\dfrac{7+Ce^{s^2}}{s^3}$

(b) Taking the inverse transform of both sides gives

$y(t)=\dfrac{7t^2}2+C\,L^{-1}\left[\dfrac{e^{s^2}}{s^3}\right]$

I don't know whether the remaining inverse transform can be resolved, but using the principle of superposition, we know that $\frac{7t^2}2$ is one solution to the original ODE.

$y(t)=\dfrac{7t^2}2\implies y'(t)=7t\implies y''(t)=7$

Substitute these into the ODE to see everything checks out:

$2\cdot7+t\cdot7t-2\cdot\dfrac{7t^2}2=14$

5 0

3 years ago