Ask questions which are clear, concise and easy to understand.
the sum of intiger and their reciprocal is 145/12. Find the intiger ?
Let the integer = x
Then reciprocal = 1/x
According to Question
x + 1/x = 145/12
=> (x2+1)/x = 145/12
=> 12x2 + 12 = 145x
=> 12x2 - 145x + 12 = 0
=> 12x2- 144x - x + 12 = 0
=> 12x(x - 12) - 1(x - 12) = 0
=> (x - 12)(12x - 1) = 0
=> x = 12 and x = 1/12
as X is integer x can't be 1/12
Therefore the required integer is 12Thanks (0)
Let the integer be x.
x + 1/x = 145/12
12x2 - 145x + 12 = 0
12x2 - 144x - x + 12 = 0
12x(x - 12) - 1(x - 12) = 0
(x - 12)(12x - 1) = 0
x = 12 and x = 1/12
Therefore the required integer is 12.Thanks (0)
how to prove the sum of first terms of ap is 42 10th term to its 30th terms is 1:3 find the first term
i think questions is : The sum of first six terms of an ap is 42 .the ratio of its 10th term is and its 30th term is 1:3. calculate the first and thirteenth term
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Let first term of AP is a and comman difference is d.
as we know
According to Question
=> 3a + 27d = a + 29d
=> 2a = 2d
=> a = d ............. (1)
Sum of first 6 terms = 42
=> a = 2
First term is 2. Common difference is 2.
Which term of AP 121,117,113.................is its first negative term
a = 121
d = -4
let an= 0
an = a + (n-1)d
=> 0 = 121 + (n-1)(-4)
=> 0 = 121 - 4n + 4
=> 4n = 125
=> n = 125/4
=> n = 31.25
as n must be natural number then n= 32
32nd term is First negative number.
a32 = 121 + (31)-4
= 121 - 124 = -3Thanks (0)
AB is a straight road leading to C, the foot of a tower, A being at a distance of 120m from C and B being 75m nearer. If the angle of elevation of the tower at B is the double of the angle of elevation of the tower at A, find the height of the tower.
Just consider the image for Reference.
Gievn : AB is straigt Road and C is the foot of tower. Let DC is tower of height h, and AC is 120cm and BC is 45 cm as it is 75cm nearer than A.
Solution : In Triangle BCD
In Triangle, ACD
From (1) and (2)
Put value of tan x in (2)
we get h = 60
so height of tower is 60cmThanks (0)
Are you sure this is the question? Please check it once.Thanks (0)
Find the probability that a non leap year chosen at random has
- 52 Sundays
- 53 Sundays
2) A non-leap year has 365 days
A year has 52 weeks. Hence there will be 52 Sundays for sure.
52 weeks = 52 x 7 = 364 days .
365– 364 = 1day extra.
In a non-leap year there will be 52 Sundays and 1day will be left.
This 1 day can be Sunday, Monday, Tuesday, Wednesday, Thursday,friday,Saturday, Sunday.
Of these total 7 outcomes, the favourable outcomes are 1.
Hence the probability of getting 53 sundays = 1 / 7.
∴ probability of getting 52 sundays = 1 - 1/ 7 = 6 / 7.Thanks (2)
The vertices of a triangle are A(-1,3), B(1,-1), C(5,1). Find the length of the median through the vertex C.
Let the points A(-1, 3), B(1, -1) and C(5, 1) be the vertices of a triangle ABC.
We know median is the line segment joining the mid point of line to opposite vertex.
Let D be the mid point of AB. Then the coordinate of D
Coorditane of D( 0,1)
Lenght of Median CD=
Using distance formulaThanks (1)
A train covers a distance of 90kms at a uniform speed. It would have taken 90 minutes less if the speed had been 15km/hr more. Calculate the original duration of the journey.
The Questions is it would have taken 30 Minutes less instead of 90 Minutes
Let “x” be the usual speed of the train
Let T1 be the time taken to cover the distance 90 km in the speed x km/hr
Let T2 be the time taken to cover the distance 90 km in the speed (x + 15) km/hr
Time = Distance/Speed
By using the given condition
T1 - T2 =
Taking 90 commonly from two fractions
Speed can't be negative so,
Speed is 45km/hThanks (1)
For what value of K are the roots of the quadratic equation kx(x-2)+6=0
For Equal roots D = 0
kx(x-2) + 6 = 0
kx2 -2kx + 6 = 0
comparing with ax2+ bx + c = 0, we get
a = k, b = -2k , c = 6
D = b2 - 4ac
Put value of a b c
(-2k)2 - 4*6*k = 0
4k2 - 24k = 0
4k = 0 or k-6=0
k = 6Thanks (0)
The height of a cone is 24 cm.A small cone is cut off at the top by a plane parallel to the base.If the volume of the remaining part be 7/8of the given cone, at what height above the base is the section made?
Let ABC be a cone of radius R, height 24cm. let this cone be cut by a place A`B` parallel to AB. O` is the center of base of the cut out cone A`B`C`. let h be the height, r be the redius of cone A`B`C`. Cleary ΔA`B`C` ~ Δ AOC
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So h/24 = r/R
h = 24r/R ..... (1)
Volume of Cone ABC = 24πR2/3 = 8πR2 ----- (2)
Remaining Portion is Know as Frustum as its volume is 7/8 of Cone ABC.
So Volume of Smaller Cone A`B`C` = 1/8 of Cone ABC
Volume of Cone A`B`C` = π r2h / 3
π r224r/3R = 8π r3/R ------ (3)
Accroding to questions
8π r3/R = 8πR2/8
8π r3/R = πR2
r3/R3 = 1/8
r/R = 1/2
put this value in (1)
We get h = 24*1/2 = 12cm
So cone is cut from the height 24-12 = 12 cmThanks (0)
The diffrence of the ages of sohrab &his father is 30 years .if the diffrence of the square of their ages is 1560 then find their ages
let age of Sohrab = x
then age of his father = x + 30
Given : (x+30)2 - (x)2 = 1560
x2 + 900 + 60x - x2 = 1560
=> 60x = 660
=> x = 11
Age of Sourab = 11
age of his father = 41Thanks (1)
The distance between the points (2,-2)&(-1,x) is 5 then one of the value of x is
Given : A (2,-2) and B(-1,x) and Distance b/w these two is 5
according to distance formula
(x2-x1)2 + (y2-y1)2 = d2
=> (-1-2)2 + (x+2)2 = 25
=> 9 + x2 + 4 +4x = 25
=> x2 + 4x - 12 = 0
=> x2 + 6x -2x - 12 = 0
=> x (x+6) -2(x+6) = 0
=> (x+6) (x-2) = 0
either x + 6 = 0 or x-2=0
so x = -6, 2Thanks (0)
Show that the points (7,3) (3,0) (0,-4) (4,-1)are vertices of rhombus
Let A(7,3) , B(3,0) , C(0,-4) and D(4,-1) represent the points.
To prove that these points are vertices of a rhombus we have to show that all sides are equal. We may calculate it using distance formula.
The diagonals are not equal.
This shows that these points are vertices of rhombus.Thanks (1)
Find the sum of all 3 digit no. Which leave a remainder 3 when divided by 5 .
First 3 Digit Number that leaves remainder 3 when divided by 5 is 103.
and last will be 998
A.P. = 103, 108, 113, ..................., 998
first term a = 103
Common Difference d = 5
nth term = 998
n = ?
as we know
nth term = a + ( n-1)d
998 = 103 + ( n-1)5
895/ 5 = n -1
179 = n -1
n = 180
now find sum of 180 terms of AP
S180 = 180/ 2 ( 103 + 998 )
=> S180 = 90 * 1001
=> S180 = 900090Thanks (0)
Solve 4x2-2(a2+b2)x+a2b2=0 find the value of x
Using Discriminant formula,
x= [2(a2+b2)+2(a2-b2)]/8 = 4a2/8 = a2/2
x= [2(a2+b2)-2(a2-b2)]/8 = 4b2/8 = b2/2Thanks (1)
If-5 is root of quadratic equation 2x^2+2px-15=0and the quadratic equation p(x^2+x)+k=0has equal roots .find the value of k
Thanks for the answer but can u answers some more questions
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as -5 is root of equation 2x2+2px-15=0
then it ll satisfy this equation, Put x = -1
2(-5)2 + 2p(-5) - 15 = 0
=> 50 - 10p -15 = 0
=> 35 = 10p
p = 3.5
And it is given that eq. p(x2+x) + k = 0 have equal roots.
px2+px + k = 0
3.5x2+3.5x + k = 0
a = 3.5, b = 3.5, c = k
so d = 0
(3.5)2 - 4 (3.5)k = 0
=> 14 k = 12.25
k = 3.0625Thanks (2)
If P(2,4) is equidistant from Q(7,0) and R (X,9), find the value of X. Also find the distance PQ.
Given: P(2,4) is equidistnat from Q(7,0) and R(X,9).
then PQ = PR
Using Distance Formula : =
PQ = =
PQ = PR
Sqauring Both Side, We get
x2 + 4 -4x + 25 = 41
=> x2 - 4x -12 = 0
=> x2 - 6x +2x -12 = 0
=> x(x-6) +2(x-6) = 0
=> (x-6) (x+2) = 0
=> x = -2 , 6Thanks (0)
PQ perpendicular OQ. The tangent to the circle with centre O at P and Q intersect at a point T. Prove that PQ and OT are right bisectors of each other.
Given: AP and AQ are tangents to the circle with centre O, AP ⊥ AQ and AP = AQ = 5 cm
we know that radius of a circle is perpendicular to the tangent at the point of contact
⇒ OP ⊥ AP and OQ ⊥ AQ
Also sum of all angles of a quadrilateral is 360° ⇒∠O + ∠P + ∠A + ∠Q = 360°
⇒∠O + 90° + 90° + 90° = 360°
⇒∠O = 360° – 270° = 90°
Thus ∠O = ∠P = ∠A = ∠Q = 90°
⇒ OPAQ is a rectangle but since adjacent sides of OPAQ i.e. AP and AQ are equal.
Thus OPAQ is a square
radius = OP = OQ = AP = AQ = 5 cm
Since diagonals of a square are perpendicular bisector of each other.
Hence PQ and OA are perpendicular bisectors of each otherThanks (0)
A 25 FOOT LADDER IS PLACED AGAINST A VERTICAL WALL OF A BUILDING.THE FOOT OF THE LADDER IS 17 FEET FROM THE BASE OF THE BUILDING,IF THE TOP OF THE LADDER SLIPS 4 FEET,THEN THE FOOT OF THE LADDER WILL SLIP.......
I think it is not 17 feet it is 7 feet.
Fiirst we need to find at what height ladder is placed.
Let BC is ladder and AC is Building with foot at C.
Now as given AB = 25
BC = 7
Using Pythagorous Theroem
(AB2) = (BC)2 + (AC)2
=> 625 = 49 + (AC)2
=> (AC)2 = 576
=> AC = 24
It slips down 4 feet it means its top is at 20 feet height
AC = 20
AB = 25
then again Using theorem
(BC)2 = 625 - 400
(BC)2 = 225
BC = 15
it mean it slips 8 feet.Thanks (0)
Sir and Madam, In our maths book, we have studied how to construct tangents from any external point without using the center of circle.I have discovered a yet new mode for the same , that is, how to construct tangents by an alternative method...
so, Can I construct the tangents by this method in Exam , If such question is asked ?