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2 years ago

5

A man sold a small cottage at a loss of 5%.Had he sold it at a gain of 7%,he would have got Rs480 more.Find the cost price of th

e cottage?

1 answer:

NARA [144]2 years ago

4 0

Answer:

Cost price of the cattage is Rs.6857.142857...

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△ABC is mapped to △A′B′C′ using the rule (x, y)→(−x, −y) followed by (x, y)→(x, −y) .

Oxana [17]

Answer: B. △ABC is congruent to △A′B′C′ because the rules represent a rotation followed by a reflection, which is a sequence of rigid motions.

Step-by-step explanation:

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3 years ago

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If you were to be driving in a car at a high amount of speed and shot a gun out the window forward (towards the windshield forwa

marin [14]

Answer:

If you shoot the bullet off the back of the car , the bullet will still be moving away from you and the gun at 1,000 mph, but now the speed of the train will subtract from the speed of the bullet. Relative to the ground, the bullet will not be moving at all, and it will drop straight to the ground.

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3 years ago

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Consider the initial value problem y′+5y=⎧⎩⎨⎪⎪0110 if 0≤t<3 if 3≤t<5 if 5≤t<[infinity],y(0)=4. y′+5y={0 if 0≤t<311 i

rosijanka [135]

It looks like the ODE is

$y'+5y=\begin{cases}0&\text{for }0\le t$

with the initial condition of $y(0)=4$ .

Rewrite the right side in terms of the unit step function,

$u(t-c)=\begin{cases}1&\text{for }t\ge c\\0&\text{for }t$

In this case, we have

$\begin{cases}0&\text{for }0\le t$

The Laplace transform of the step function is easy to compute:

$\displaystyle\int_0^\infty u(t-c)e^{-st}\,\mathrm dt=\int_c^\infty e^{-st}\,\mathrm dt=\frac{e^{-cs}}s$

So, taking the Laplace transform of both sides of the ODE, we get

$sY(s)-y(0)+5Y(s)=\dfrac{e^{-3s}-e^{-5s}}s$

Solve for $Y(s)$ :

$(s+5)Y(s)-4=\dfrac{e^{-3s}-e^{-5s}}s\implies Y(s)=\dfrac{e^{-3s}-e^{-5s}}{s(s+5)}+\dfrac4{s+5}$

We can split the first term into partial fractions:

$\dfrac1{s(s+5)}=\dfrac as+\dfrac b{s+5}\implies1=a(s+5)+bs$

If $s=0$ , then $1=5a\implies a=\frac15$ .

If $s=-5$ , then $1=-5b\implies b=-\frac15$ .

$\implies Y(s)=\dfrac{e^{-3s}-e^{-5s}}5\left(\frac1s-\frac1{s+5}\right)+\dfrac4{s+5}$

$\implies Y(s)=\dfrac15\left(\dfrac{e^{-3s}}s-\dfrac{e^{-3s}}{s+5}-\dfrac{e^{-5s}}s+\dfrac{e^{-5s}}{s+5}\right)+\dfrac4{s+5}$

Take the inverse transform of both sides, recalling that

$Y(s)=e^{-cs}F(s)\implies y(t)=u(t-c)f(t-c)$

where $F(s)$ is the Laplace transform of the function $f(t)$ . We have

$F(s)=\dfrac1s\implies f(t)=1$

$F(s)=\dfrac1{s+5}\implies f(t)=e^{-5t}$

We then end up with

$y(t)=\dfrac{u(t-3)(1-e^{-5t})-u(t-5)(1-e^{-5t})}5+5e^{-5t}$

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4 years ago

The graph of f(x) = 2x + 3 shifts 10 units to the right when it is replaced with the graph of f(x) = 2x − k. What is the value o

Nezavi [6.7K]

Answer:

k=17

Step-by-step explanation:

Given F(x)=2x+3 shifts 10 units when it is replaced with the graph f(x)=2x-k.

The method to solve this kind of questions is find where the graph meets the x axis then add ten units to it .The obtained x coordinate is the point where the second graph meets the x axis.

Now first find where the F(x)=2x+3 meets the x axis by substituting F(x)=0.

$2x+3=0$

$x=-\frac{3}{2}$

Now add 10 to this x coordinate , we get x1= $10-3/2$ = $17/2$

this is x coordinate where the second line meets the x-axis.

The second line is F(x)=2x-k.

now f(x)=0 and x=17/2

On substitution we get $0=2*17/2-k$

k=17

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3 years ago

Find each percent of change. Round answers to the nearest tenth of a percent is necessary. 76 is decreased to 55

Allushta [10]

76 -55= 21
76 is decreased by 21.

We initially start with 76, so now we have to divide 21 by 76:
21/76* 100%= 27.63%

Rounded to the nearest tenth, the final answer is 27.6%~

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3 years ago