What is the range of values for measurements 300 +2%

A store had 150 laptops in the month of January. Every month, 20% of the laptops were sold and 10 new laptops were stocked in th

Softa [21]

<u>Answer-</u>

$\boxed{\boxed{f(n)=0.8(f(n-1))+10,\ f(0)=150,\ \ n>0}}$

<u>Solution-</u>

Here, n represents the number of months and f(n) represents the number of laptops in the store after n months.

As the store had 150 laptops in the month of January or at the beginning.

So $f(0)=150$

Every month, 20% of the laptops were sold and 10 new laptops were stocked in the store.

As 20% of laptops were sold, so 80% were in the store.

So, after one month total number of laptops in the store,

$\Rightarrow f(1)=0.8(150)+10$

$\Rightarrow f(1)=0.8(f(0))+10$ $(\because f(0)=150)$

Again after one month total number of laptops in the store,

$\Rightarrow f(2)=0.8\times (0.8(150)+10)+10$

$\Rightarrow f(2)=0.8(f(1))+10$ $(\because f(1)=0.8(150)+10)$

Analyzing the pattern, the recursive function f(n) will be,

$f(n)=0.8(f(n-1))+10,\ f(0)=150,\ \ n>0$

3 0

2 years ago

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Nikolay [14]

Replace x with π/2 - x to get the equivalent integral

$\displaystyle \int_{-\frac\pi2}^{\frac\pi2} \cos(\cot(x) - \tan(x)) \, dx$

but the integrand is even, so this is really just

$\displaystyle 2 \int_0^{\frac\pi2} \cos(\cot(x) - \tan(x)) \, dx$

Substitute x = 1/2 arccot(u/2), which transforms the integral to

$\displaystyle 2 \int_{-\infty}^\infty \frac{\cos(u)}{u^2+4} \, du$

There are lots of ways to compute this. What I did was to consider the complex contour integral

$\displaystyle \int_\gamma \frac{e^{iz}}{z^2+4} \, dz$

where γ is a semicircle in the complex plane with its diameter joining (-R, 0) and (R, 0) on the real axis. A bound for the integral over the arc of the circle is estimated to be

$\displaystyle \left|\int_{z=Re^{i0}}^{z=Re^{i\pi}} f(z) \, dz\right| \le \frac{\pi R}{|R^2-4|}$

which vanishes as R goes to ∞. Then by the residue theorem, we have in the limit

$\displaystyle \int_{-\infty}^\infty \frac{\cos(x)}{x^2+4} \, dx = 2\pi i {} \mathrm{Res}\left(\frac{e^{iz}}{z^2+4},z=2i\right) = \frac\pi{2e^2}$

and it follows that

$\displaystyle \int_0^\pi \cos(\cot(x)-\tan(x)) \, dx = \boxed{\frac\pi{e^2}}$

7 0

1 year ago

How to find the answers ?

maria [59]

AD || BC and BD is the transversal,

Therefore, angle DBC = angle ADB = 42° [alternate angles]

AD || BC and BD is the transversal,

angle BAD + angle ABD + angle DBC = 180° [co-interior angles]

or, 106°+ angle ABD + 42° = 180°

or, 148° + angle ABD = 180°

or, angle ABD = 180°-148° = 32°

Therefore, angle ABC = (32+42)° = 74°

AB||CD and BC is the transversal,

angle ABC + angle BCD = 180° [co-interior angles]

or, 72+2x+12 = 180

or, 84+2x = 180

or, 2x=180-84 = 96

or, x = 48

Answers: a)48°

b)42°

c)72°

3 0

2 years ago

Equation for ( 6, 13) ( 7900, 5 )

tia_tia [17]

For this case we have that by definition, the equation of the line in the slope-intersection form is given by:

$y=mx+b$

Where:

m: It is the slope of the line

b: It is the cut-off point with the y axis

We have the following points through which the line passes:

$(x_ {1}, y_ {1}) :( 6,13)\\(x_ {2}, y_ {2}): (7900,5)$

We find the slope of the line:

$m = \frac {y_ {2} -y_ {1}} {x_ {2} -x_ {1}} = \frac {5-13} {7900-6} = \frac {-8} {7894} = - \frac {4} {3947}$

Thus, the equation of the line is of the form:

$y = - \frac {4} {3947} x + b$

We substitute one of the points and find b:

$13 = - \frac {4} {3947} (6) + b\\13 = - \frac {24} {3947} + b\\13+ \frac {24} {3947} = b\\b = \frac {51335} {3947}$

Finally, the equation is:

$y = - \frac {4} {3947} x + \frac {51335} {3947}$

Answer:

$y = - \frac {4} {3947} x + \frac {51335} {3947}$

5 0

3 years ago

Could you help me with this for 50? points.

ruslelena [56]

+1,-1 and your other would be 1,1

3 0

1 year ago

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