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2 years ago

The sum of two numbers is 196. One number is 64 more than the other. Find the numbers.

Mathematics

2 answers:

anzhelika [568]2 years ago

6 0

Answer:

Step-by-step explanation:

Hi there!

Let one of the number be "x" then the next number is x+64.

According to the question,

The sum of two numbers is 196.

i.e x + (x+64) = 196

or, 2x +64 = 196

Subtract 64 from both sides,

2x +64-64 = 196 - 64

2x = 132

Divide both sides by 2,

$\frac{2x}{2}$ = $\frac{132}{2}$

x = 66

And x+64 = 66+64 = 130

Therefore, the one number was 66 and another was 130.

Hope it helps!

ValentinkaMS [17]2 years ago

4 0

Answer:

I just guessed random numbers until I got 66 and 130.

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I really need a answer ASAP

Elza [17]

Answer:

e and c

Step-by-step explanation:

4 0

3 years ago

I need help with number 3 please and thank you

murzikaleks [220]

So here we're dealing with equivalent fractions.
It's really simple to find the answer, so I'll try to explain the best I can.

2 dogs / 5 cats is really just 2/5.
If we want to find an equivalent fraction, we have to multiply the numerator and the denominator by the same number.

Currently the number of cats is 5, and we need it to be 140. What we need to do is find the number it has to be multiplied by to equal 140, which is 140 divided by 5. 140 divided by 5 is 28, so 5 x 28 = 140!

We need to multiply the denominator (5) by 28, so that we can get 140. What we have now is ?/140.

Like I said, to find an equivalent fraction, we need to multiply the numerator by the same number as we did the denominator, which is 28!

2 x 28 = 56.

So 2/5 is the same as 56/140.

The answer is D) 56 Dogs/140 Cats.

Hope this helps!
If you're confused about anything leave me a reply and I'll try to explain the best that I can!

3 0

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

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