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Ask QuestionPosted by Anish Kadam 5 years, 3 months ago
- 3 answers
Posted by Tejas Patil 5 years, 3 months ago
- 1 answers
Yogita Ingle 5 years, 3 months ago
Let the present age of father = x years and the present age of son = y years
According to question:
x + y = 45 ...(1)
Again, 5 years ago, the product of their ages was 124, therefore,
(Age of man 5 years ago) × (Age of son 5 years ago) = 124
(x – 5) × (y – 5) = 124
⇒ (45 – y - 5) ×(y – 5) = 124 [Using (1)]
⇒ 40y – y2 - 200 + 5y = 124
⇒ y2 - 45y + 324 = 0
⇒ y2 - 36y - 9y + 324 = 0
⇒ y(y - 36) -9 (y - 36) = 0
⇒ (y - 36)(y - 9) = 0
⇒ y = 36 or y = 9
For y = 36, x = 45 - 36 = 9 which is not possible.
For y = 9, x = 45 - 9 = 36
Therefore, the present age of father = 36 years and the present age of son = 9 years.
Posted by Abhay Singh 5 years, 3 months ago
- 1 answers
Gaurav Seth 5 years, 3 months ago
When a, b, c are real numbers, a 0: If = b² -4 a c = 0, then roots are equal (and real). If = b² -4 a c > 0, then roots are real and unequal. If = b² -4 a c < 0, then roots are complex.
Posted by Ayush Jena 5 years, 3 months ago
- 0 answers
Posted by Purab Saini 5 years, 3 months ago
- 5 answers
Pranav K.P 5 years, 3 months ago
₷Āńïk₳ Dĥãpãtè ₯ 5 years, 3 months ago
Posted by Sachin Kohli 5 years, 3 months ago
- 0 answers
Posted by Mega K 5 years, 3 months ago
- 3 answers
Mandaloju Bharath 5 years, 3 months ago
₷Āńïk₳ Dĥãpãtè ₯ 5 years, 3 months ago
Yogita Ingle 5 years, 3 months ago
Put x = 2
x2-kx+10 = 0
(2)2 - k(2) + 10 = 0
4 - 2k + 10 = 0
-2x +14 = 0
-2x = -14
x = 14/2 = 7
Posted by Sarjit Rathod 5 years, 3 months ago
- 1 answers
₷Āńïk₳ Dĥãpãtè ₯ 5 years, 3 months ago
Posted by Tanishq Rai 5 years, 3 months ago
- 0 answers
Posted by Pranav K.P 4 years, 6 months ago
- 1 answers
Sia ? 4 years, 6 months ago
Let A (x1, y1) and B (x2, y2) be the endpoints of the given line segment AB and C(x, y) be the point which divides AB in the ratio m : n. We want to find the coordinates (x, y) of C. Now draw perpendiculars of A, C, B parallel to Y coordinate joining at P, Q, and R on X-axis.
Posted by ?☺???Hrishi Khapekar???☺? 5 years, 3 months ago
- 2 answers
Rohit Kumar 5 years, 3 months ago
Posted by Prajnasree Behera 5 years, 3 months ago
- 3 answers
Prajnasree Behera 5 years, 3 months ago
Sathwik Nannaware 5 years, 3 months ago
Posted by Madhiha ??? 5 years, 3 months ago
- 1 answers
Gaurav Seth 5 years, 3 months ago
from circumcentre distance of every vertex is same
let circumcentre point C
so AC=BC=CO
AC^2=BC^2=CO^2
by distance formula
AC^2 = BC^2
[(x-a)^2+(y-0)^2] = [(x-0)^2+(y-b)^2]
by solving
x^2+a^2 -2ax+ y^2 = x^2 +y^2+ b^2-2yb
2(by-ax)= b^2 -a^2 ---- (1)
NOW AC^2=OC^2
(x-0)^2 + (y-b)^2 = x^2 + y^2
by solving this equation y=b/2
put the value of y in (1) and we will get x=a/2
So answer is (a/2,b/2)
Posted by Siddhartha Mishra 5 years, 3 months ago
- 0 answers
Posted by Divya Divya 5 years, 3 months ago
- 4 answers
Posted by Anchal Sheoran 5 years, 3 months ago
- 3 answers
Dhanush N 5 years, 3 months ago
Posted by Puneet Sharma 5 years, 3 months ago
- 1 answers
Yogita Ingle 5 years, 3 months ago
141x+93 y =189 .........(i)
93 x + 141 y=45.........(ii)
Add (i) and (ii), we get
141x+93 y =189
93 x + 141 y=45
_______________
234x + 234 y= 234
x + y=1........ (iii)
Subtract (i) and (ii), we get
141x+93 y =189
93 x + 141 y=45
_______________
48 x - 48 y=144
x - y = 3........ (iv)
Adding (iii) and (iv), we get
x + y=1
+ x - y = 3
____________
2x = 4
x = 2
Put x = 2 in (iii)
2 + y = 1
y = 1 - 2 = -1
Posted by Bhagyashri Marwadi 5 years, 3 months ago
- 3 answers
Posted by Ritik Ranjan 5 years, 3 months ago
- 2 answers
Gaurav Seth 5 years, 3 months ago
For every quadratic equation, there can be one or more than one solution. These are called the roots of the quadratic equation.
For a quadratic equation ax2+bx+c = 0,
the sum of its roots = –b/a and the product of its roots = c/a.
Posted by Khushi Sambharia 5 years, 3 months ago
- 4 answers
Posted by Manav Sandhu 5 years, 3 months ago
- 4 answers
Sri Anish 5 years, 3 months ago
Anirudh Jaat From Haryana 5 years, 3 months ago
Anupama ?? 5 years, 3 months ago
Posted by Chhavi Sharma 5 years, 3 months ago
- 1 answers
Gaurav Seth 5 years, 3 months ago
Q u e s t i o n : Find the area of a triangle ABC with A(1, - 4) and mid-points of sides through A being (2, - 1) and (0, - 1)
A n s w e r :
Let E be the mid point of AB
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Posted by Mahi Lodee 5 years, 3 months ago
- 2 answers
Dev Sangwan 5 years, 3 months ago
Posted by Rahul Yadav 5 years, 3 months ago
- 2 answers
Posted by Mauryan Princess 5 years, 3 months ago
- 3 answers
Dhanush N 5 years, 3 months ago
Gaurav Seth 5 years, 3 months ago
Step - by - step explanation :
a =5, d =5, tn =995
995 = 5 + (n-1) × 5
995 = 5 + 5n - 5
995 = 5 n
n = 995 ÷ 5 = 199

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Anupama ?? 5 years, 3 months ago
0Thank You