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3 years ago

What conclusion can be based on the given statement?

Mathematics

2 answers:

Inessa [10]3 years ago

8 0

Answer:

The option B. q = 43.0 is correct

Step-by-step explanation:

Whole Number : It is defined as non-negative integer whose fraction or decimal part is always 0.

Now, it is given that q is a whole number between 42.3 and 43.1

So, to find q :

The given options are :

A. q = 42.1 : As the decimal part is not 0 in 42.1 so it cannot be whole number. So this option is rejected.

C. q = 43.4 : As the decimal part is not 0 in 43.4 so it cannot be whole number. So this option is rejected.

B. q = 43.0 : The decimal part in 43.0 is 0 so it is an whole number.

Also 42.3 < q = 43.0 < 43.1

So, the required value of q = 43.0

Hence, The option B. q = 43.0 is correct

Mariulka [41]3 years ago

7 0

42.3 < 43 < 43.1
43 is inbetween the two values, so it can be q.

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3 years ago

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3 years ago

Consider the following. (A computer algebra system is recommended.) y'' + 3y' = 2t4 + t2e−3t + sin 3t (a) Determine a suitable f

drek231 [11]

First look for the fundamental solutions by solving the homogeneous version of the ODE:

$y''+3y'=0$

The characteristic equation is

$r^2+3r=r(r+3)=0$

with roots $r=0$ and $r=-3$ , giving the two solutions $C_1$ and $C_2e^{-3t}$ .

For the non-homogeneous version, you can exploit the superposition principle and consider one term from the right side at a time.

$y''+3y'=2t^4$

Assume the ansatz solution,

${y_p}=at^5+bt^4+ct^3+dt^2+et$

$\implies {y_p}'=5at^4+4bt^3+3ct^2+2dt+e$

$\implies {y_p}''=20at^3+12bt^2+6ct+2d$

(You could include a constant term <em>f</em> here, but it would get absorbed by the first solution $C_1$ anyway.)

Substitute these into the ODE:

$(20at^3+12bt^2+6ct+2d)+3(5at^4+4bt^3+3ct^2+2dt+e)=2t^4$

$15at^4+(20a+12b)t^3+(12b+9c)t^2+(6c+6d)t+(2d+e)=2t^4$

$\implies\begin{cases}15a=2\\20a+12b=0\\12b+9c=0\\6c+6d=0\\2d+e=0\end{cases}\implies a=\dfrac2{15},b=-\dfrac29,c=\dfrac8{27},d=-\dfrac8{27},e=\dfrac{16}{81}$

$y''+3y'=t^2e^{-3t}$

$e^{-3t}$ is already accounted for, so assume an ansatz of the form

$y_p=(at^3+bt^2+ct)e^{-3t}$

$\implies {y_p}'=(-3at^3+(3a-3b)t^2+(2b-3c)t+c)e^{-3t}$

$\implies {y_p}''=(9at^3+(9b-18a)t^2+(9c-12b+6a)t+2b-6c)e^{-3t}$

Substitute into the ODE:

$(9at^3+(9b-18a)t^2+(9c-12b+6a)t+2b-6c)e^{-3t}+3(-3at^3+(3a-3b)t^2+(2b-3c)t+c)e^{-3t}=t^2e^{-3t}$

$9at^3+(9b-18a)t^2+(9c-12b+6a)t+2b-6c-9at^3+(9a-9b)t^2+(6b-9c)t+3c=t^2$

$-9at^2+(6a-6b)t+2b-3c=t^2$

$\implies\begin{cases}-9a=1\\6a-6b=0\\2b-3c=0\end{cases}\implies a=-\dfrac19,b=-\dfrac19,c=-\dfrac2{27}$

$y''+3y'=\sin(3t)$

Assume an ansatz solution

$y_p=a\sin(3t)+b\cos(3t)$

$\implies {y_p}'=3a\cos(3t)-3b\sin(3t)$

$\implies {y_p}''=-9a\sin(3t)-9b\cos(3t)$

Substitute into the ODE:

$(-9a\sin(3t)-9b\cos(3t))+3(3a\cos(3t)-3b\sin(3t))=\sin(3t)$

$(-9a-9b)\sin(3t)+(9a-9b)\cos(3t)=\sin(3t)$

$\implies\begin{cases}-9a-9b=1\\9a-9b=0\end{cases}\implies a=-\dfrac1{18},b=-\dfrac1{18}$

So, the general solution of the original ODE is

$y(t)=\dfrac{54t^5 - 90t^4 + 120t^3 - 120t^2 + 80t}{405}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-\dfrac{3t^3+3t^2+2t}{27}e^{-3t}-\dfrac{\sin(3t)+\cos(3t)}{18}$

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3 years ago

What is the total cost of 0.5 pound of peaches selling for $0.80 per pound and 0.7 pound of oranges selling for $0.90 per pound?

swat32

Given situation : 0.5 pound of peaches selling for 0.80 dollars/ pound 0.7 pound of oranges selling for 0.90 dollars / pound. Solution Given number 1 : Peaches => 0.5 pound = meaning, ½ pound is available, And 1 pound of it costs 0.80 dollars. Let’s solve: => 0.80 dollars * 0.5 => 0.40 dollars – the price of the peaches. Given number 2 : Oranges => 0.7 pounds of oranges, meaning less than 1 pound. And 1 pound costs 0.90 dollars Let’s solve to get the anwer => 0.7 * .90 => 0.63 dollars – the costs of 0.7 pounds of oranges,

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3 years ago