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3 years ago

5

A store has a 40\%40%40, percent off sale on headphones. With this discount, the price of one pair of headphones is \$36$36dolla

r sign, 36. What is the original price of the pair of headphones?

2 answers:

zloy xaker [14]3 years ago

8 0

Let
X=<span>the original price of the pair of headphones
then
if the % </span>off sale is 40--------------- > (1-0.40)*X=<span>$36
x=36/0.60=60

the answer is
</span>the original price of the pair of headphones was $60

vlabodo [156]3 years ago

3 0

Let x be the original price of headphones.

Since there is 40% off, the price after discount will be 60% of x. So we can set this information as:

$60 \text { percent of x} =36$

$\frac{60}{100} \cdot x=36$

$x=36\cdot\frac{100}{60}$

$x=36\cdot\frac{10}{6}$

$x=6\cdot10=60$

Therefore, original price of headphones is $60.

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Answer:

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Step-by-step explanation:

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3 years ago

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3 years ago

Read 2 more answers

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

One hundred tickets, numbered 1, 2, 3, …, 100, are sold to 100 people for a drawing. Four different prizes are awarded, includin

stira [4]

Solution :

It is given that four different prizes were awarded. So,

a). 4 ways for person 47 to win a prize

99 ways to give out the 2nd prize

98 ways to give the 3rd prize

97 ways to give the last prize

∴ P(99,3) = 99 x 98 x 97

b). 1 way to give person 47 their prize

1 way to give person 19 their prize

98 ways to give out the 3rd prize

97 ways to give out the last prize

So, P(98,2) = 98 x 97

8 0

3 years ago

Consider an angle with a measure of θ=225∘.

wel

Answer:

cos 225° = $-\frac{\sqrt{2} }{2}$

sin 225° = $-\frac{\sqrt{2} }{2}$

tan 225° = 1

Step-by-step explanation:

I cannot sketch a diagram, but a 225° angle is a 3rd quadrant angle and the reference angle is 45° (225 - 180 = 45)

cos and sin are negative in the 3rd quadrant and the tan is positive

cos 225° = - cos 45° = $-\frac{\sqrt{2} }{2}$

sin 225° = -sin 45° = $-\frac{\sqrt{2} }{2}$

tan 225° = tan 45° = 1

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2 years ago