${0\le x\lt \frac{5}{3}} or, x\gt 10$
Here, p = 3x$^{2}$-35x+50
Now. if the company makes a profit
p >0
=> 35$^{2}$ + 50 > o
=> 3x$^{2}$ -30x – 5x + 50 > 0
=> 3x(x- 10) – 5(x – l0) >0………………………………….(i)
As this equation (i) is greater than 0. so the value of the two roots must have different values in different intervals.
Now . front the equation (i), we have
The value of x is greater than 10 . So, x > 10
0r. the value of x is less than $\frac{5}{3}$ and greater tlum or equal to 0 i.e. 0 $\le x\lt \frac{5}{3}$
because advertising cost can not be negative.
So. if the company makes a prof it. the values of x is ${0\le x\lt \frac{5}{3}} or, x\gt 10$
37.5 meter.
576 Tk. would he saved when the breadth 3 meters less
1 Tk. would be saved when the breadth is $\frac{3}{576}$ meters less
7.300 Tk. would be saved when the breadth Is $\frac{3\times 7200}{576}$ = 37.5 meter less
227 & 353
Since the sum of the 3-digits is 11 and each digit represents a prime number,
so the 3-digits may be 2,2,7 or 3. 3. 5 because 2 + 2 + 7 = 11 and 3 + 3 + 5 = 11
Now using the digit 2. 2. 7 we have the prime number 227. Because other two numbers i,e 722 and 272 are divisible by 2 anti thus are not prime.
Again. using the digit 3, 3, 5 we have the number 353 which is a prime number. But 533 which
is divisible by l3 and thus 533 is not prime and 335 is divisible by 5 and thus 335 is not prime
So. the three-digit prime number whose sum of the digits is 11 and each digit representin a
prime number is 227 & 353. (ans)
See Explanation
Let, $\frac{a}{q-r}=\frac{b}{r-p}=\frac{c}{p-q}$ = k
so, $\frac{a}{q-r}$ = k
so,
a=k(q- r); b =k(r-P) & C=k(P-q)
L.H.S = a+b+c
= k(q-r) + k(r-p) + k(P-q)
=k(q-r+ r-p +P-q)
=k $\times$ 0
=0
R.H.S. =pa + qb + rc
=p[k(q – r)] + q [k(r-p)] + r [k(p- q)]
= p(kq – kr) + q (kr – kq) + r(kp – kq)
= pkq – pkr + qkr — pkq + pkr – qkr
= 0
Therefore. a + b + c =pa + qb + rc
$\angle A = \frac{180^{\circ }}{2}=90^{\circ }$
Given. Let ABCD be a cyclic parallelogram
To prove. ABCD is a rectangle
We will proof. a rectangle is a parallelogram with one angle 90°
So. we have to prove angle $\angle$A = 90°
Since ABCD is a parallelogram ,
$\angle$A = $\angle$C ……….(i) [Oppsite angles of parallelogram are equal]
In cyclic parallelogram ABCD
$\angle A+\angle C$ = 180°
=> $\angle A+\angle A$ = 180°
=> 2$\angle$A = 180°
=> $\angle A = \frac{180°}{2}=90°$
He traveled at a speed of 9 km/hr.
Let. the speed be x km/hr.
Since distance is 108 km, time $\frac{108}{x}$ hrs
When speed is increased by 3 km/hr, speed is = (x + 3) km/hr
so, The required time = $\frac{108}{x+3}$ hrs
According t0 the question,
$\frac{108}{x}$-$\frac{108}{x+3}$ = 3
=> $\frac{108(x+3)-108x}{x(x+3)} =3$
=> 324 = 3x$^{2}$+9x
=> -3x$^{2}$ -9x+324 = 0
=> -3(x$^{2}$+3x-108) = 0
=> x$^{2}$+12x-9x-108 = 0
=> (x-9)(x+12) = 0
=> So either, x -9 = 0 Or, r + 12 = 0
If, x-9 =0 then x=9
If x + l2 = 0 then x =- 12
so, The speed 9km [Since velocity cant not be negative]
He traveled at a speed of 9 km/hr. (ans)
(X. y) = (6, 1)
$\frac{x}{2}+\frac{6}{y}$=9
=> $\frac{xy+12}{2y}$=9
so, xy +12 = 18y…………….(i)
$\frac{x}{3}+\frac{2}{y}=4$
=> $\frac{xy+6}{3y}$ =4
so, xy + 6 = 12y (ii)
Now, [(i)- (ii)] we have
xy + 12 =18y
xy + 6 = 12y [বিয়গ করা হল]
6 = 6y
putting the value of yin (ii) we have
=> x= 12-6
SO, x=6
(x,y) = (6,1)
