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find the prime factorisation of the denominator of the rational number equivalent to 8.39

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Name shape of the curve of a quadratic polinomial

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The graph of a quadratic function is called a Parabola.
A parabola is roughly shaped like the letter "U".
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A Boat takes 3 hours to go 45 km downstream and IT returns in 9 hours find the speed of stream and that a boat in still water

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Ans. Let Speed of boat in still water = x km/h
Speed of stream = y km/h
According to question,
Case 1 Downstream :
Time taken = 3 hours
total distance = 45 km
Speed of boat downstream = x + y
So, {tex}{45\over 3} = x+y{/tex}
=> x+y = 15 ............. (1)
Case 2 Upstream :
Time taken = 9 hourstotal distance = 45 km
Speed of boat upstream = x  y
So, {tex}{45\over 9} = xy{/tex}
=> x  y = 5 ..............(2)
Adding (1) and (2) we get
=> 2x = 20
x = 10
From (2). 10  y = 5
=> y = 5
So speed of boat in still water = 10 Km/h
Speed of stream = 5 Km/h
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GIVE EXAMPLE OF POLYNOMIALS p(x),g(x)and r(x), which satisfy the division algorithm and:
 deg p(x)=deg q(x)
 deg q(x)=deg r(x)
 deg r(x)=0
PLS GIVE THE ANSWER........

THANK YOU VERY MUCH.........
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We can write many such examples
(i) deg p(x) = deg q(x)
P(x) = x^{2}, q(x) = x^{2}, g(x) = 1, R(x) = 0
(ii) deg q(x) = deg r(x)
q(x) = x , R(x) = x , p(x) = x^{2} + x, g(x) = x^{2}
(iii) deg r(x) = 0
q(x) = 1 , R(x) = 1 , p(x) = x+1, g(x) = x
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Let a b c p be rational numbers such that p is not perfect cube . if a +bp^1/3+cp^2/3=0 then prove a=b=c=0

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As p is not a perfect cube. Hence p^{1/3 }and p^{1/3 }, p^{2/3 }are irrartional numbers. Then we have b p^{1/3 }+c p^{2/3 }as irrational and a as rational. As sum of a rational number and an irrational number cannot give us a zero as answer so all the coefficiants (a,b and c) must be zero in general.
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If a+b=5 and 3a+2b=20, Then value of a+b is:
1) 10
2) 15
3) 20
4) 25

you have already given value of a+b = 5 then what is the use of asking.
it should be a printing mistake. you have to find value of 3a+b
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Ooo.
But this ques. had come in chs exam
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Question Must be : If a + b = 5 and 3a + 2b = 20, then 3a + b will be
Ans. 4) 25
a + b = 5 .............(1)
3a + 2b = 20 .........(2)
From (1), a = 5  b
put this value in (2), we get
=> 3( 5  b ) + 2b = 20
=> 15  3b + 2b = 20
=>  b = 5
=> b = 5
put values of b in (1), we get
=> a  5 = 5
=> a = 10
Now, 3a + b = 3(10 )  5
= 30  5 = 25
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If cos (a+b)=0, then sin (ab)=
1) cos 2b
2) cos b
3) sin 2a
4) sin a

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1) cos 2b
Explanation:
cos (a+b) =0
=> cos (a+b) = cos 90
=> a+b = 90
=> a = 90 b
Now
sin(a+b)
= sin(90bb)
= sin(902b)
= cos 2b [as sin (90  x) = cos x]
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According to trigonometric ratios of 30 degree and 60 degree, in an isosceles triangle ABC in which D is mid point of side BC, the values of side AB and AC is 2a and BD and CD is a why??

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The value of
cot x
<hr/>cot xcot 3x
+
tan x
<hr/>tan x tan 3x
Will be??

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In an isosceles triangle ABC, if AB=BC and AB^{2}=2AC^{2} ,then angle C will be???

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Ans.
Given: AC = BC
ABC is an Isosceles triangleAB^{2} = 2 AC^{2}
To Find = {tex}\angle C{/tex}
Proof :
As AB^{2} = 2AC^{2}
=>^{ }AB^{2} = AC^{2 } + AC^{2}
=> ^{ }AB^{2} = AC^{2 } + BC^{2} [Given AC = BC]
So By Converse of Pythagoras theorem
{tex}\angle C = 90^o{/tex}
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SecA (1+SecA)(SecAtanA)=1

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secA(1sinA)(secA+tanA)
=(secAtanA)(secA+tanA)
=1... using identity
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Find the zero of the quadratic polynomials and verify the relationship between the zero and the coefficient :
x² 2x 8

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Ans. To find zeros, we ll equate it with 0
=> x^{2 } 2x  8 = 0
=> x^{2}  4x + 2x  8 = 0
=> x(x4) + 2( x4) = 0
=> (x  4) ( x + 2 ) = 0
=> x  4 = 0 Or x + 2 = 0
=> x = 4, 2
Zeros of the polynomial are 4 , 2
On comparing with ax^{2} + bx + c, we get a = 1, b = 2 and c = 8
We know Sum of zeroes = {tex}{b\over a}{/tex}
=> 4 + (2) = {tex} {2\over 1}{/tex}
=> 2 = 2
Again Product of zeroes = {tex}c\over a{/tex}
=> {tex}4 \times (2) = {8\over 1}{/tex}
=> 8 = 8
Hence verified
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How to get full marks in maths exam 201718

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Unit tests of class 10 pattern 2017

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Every school has to take three periodic tests for every subject and the best two will be considered.
Periodic tests has a weightage of 10 marks.
23 chapters will be assessed in every subject.
All the best. Make the most of your summer vacations!
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What is best way of solving out an root term?

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what are properties of integers

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Ans.
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What is procedure of using elimination method?

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Ans. The elimination method of solving systems of equations is also called the addition method. To solve a system of equations by elimination we transform the system such that one variable "cancels out".
Here are the steps to follow:
Step 1 : Try to eliminate a variable as you add the left sides and the right sides of the two equationsStep 2 : Set the sum resulting from adding the left sides equal to the sum resulting from adding the right sides
Step 3 : Solve for the variable that was not cancelled or eliminated
Step 4 : Use the answer found in step 3 to solve for the other variable by substituting this value in one of the two equations
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tan²A+cot²A+2=sec²A cosec²A

i forgot to add 2

question is wrong . it should be tan^{2}a+cot^{2}a = sec^{2}a+cosec^{2}a
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tan^{2}a + cot^{2}a + 2
= sec^{2}a  1 + cosec^{2}a  1 + 2
= sec^{2}a  2+2 + cosec^{2}a
=sec^{2}a + cosec^{2}a
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When to do LCM in problem sums?

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When you have to find smallest least minimum number
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When to do HCF in problem sums?

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When you need to find out maximum or largest value
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What are the zeroes of quadratic polynomial x^{2}+99x+127?

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We need to find zeroes of x^2+99*x+127=0
We have certainly got a formula for determining the zeroes of a 2 degree equation.That is called Discriminant formula or D formula or Shri Dharacharya Method.
Roots of the 2 degree equation are given by=(b+(b^24*a*c)^(1/2))/2*a and (b(b^24*a*c)^(1/2))/2*a
So, Zeroes of this equation is given by
(99+(99^24*1*127)^(1/2))/2*1 and (99(99^24*1*127)^(1/2))/2*1
Hence,zeroes are 1.3 and 97.7 respectively.Thanks (0)
If the product of zeroes of the quadratic polynomial f (x) =x square 4x+k is 3 . Find the value of k

Please answer me fast
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Product of the zeroes =c/a
=k/1=3
=k=3
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Ans. x^{2} 4x + k
on comparing with ax^{2 } + bx + c we get
a= 1 b = 4 and c = k
we know that
product of roots = {tex}c\over a{/tex}
=> {tex}3 = {k\over 1}{/tex}
=> k = 3
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Write the quadratic polynomial having 1/4,1 as it's zeroes

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Ans. Let equation is ax^{2} + bx + c = 0, roots are {tex}{1\over 4},1{/tex}.
We know that,
Sum of roots = {tex}{b\over a}{/tex}
=> {tex}{1\over 4} + 1 = {b\over a}{/tex}
{tex}=> {3\over 4} = {b\over a}{/tex}
=> b = 3 and a = 4
Also
Product of roots = {tex}c\over a{/tex}
=> {tex}{1\over 4}\times 1 = {c\over a}{/tex}
{tex}=> {1\over 4}= {c\over a}{/tex}
c = 1 and a = 4
Put Values of a,b,c in equation we get,
4x^{2}  3x 1 = 0 Requried equation
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the area of isoceles triangle EFG is 60 sq .m and EF=EG=13 m find the FG

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Let the length of FG be x m.
Using Heron's formula,
s = {tex}{{13 + 13 + x} \over 2} = {{26 + x} \over 2}{/tex}
Area of triangle EFG = 60 sq.m
=> {tex}\sqrt {\left( {{{26 + x} \over 2}} \right)\left( {{{26 + x} \over 2}  13} \right)\left( {{{26 + x} \over 2}  13} \right)\left( {{{26 + x} \over 2}  x} \right)} {/tex} = 60
=> {tex}\sqrt {\left( {{{26 + x} \over 2}} \right)\left( {{{26 + x  26} \over 2}} \right)\left( {{{26 + x  26} \over 2}} \right)\left( {{{26 + x  2x} \over 2}} \right)} {/tex} = 60
=> {tex}\sqrt {\left( {{{26 + x} \over 2}} \right)\left( {{x \over 2}} \right)\left( {{x \over 2}} \right)\left( {{{26  x} \over 2}} \right)} {/tex} = 60
=> {tex}{x \over 2}\sqrt {\left( {{{676  {x^2}} \over 4}} \right)} = 60{/tex}
=> {tex}{x \over 4}\sqrt {676  {x^2}} = 60{/tex}
=> {tex}x\sqrt {676  {x^2}} = 240{/tex}
Squaring both sides,
=> {tex}{x^2}\left( {676  {x^2}} \right) = 57600{/tex}
=> {tex}{x^4}  676{x^2} + 57600 = 0{/tex}
=> {tex}{x^4}  576{x^2}  100{x^2} + 57600 = 0{/tex}
=> {tex}{x^2}\left( {{x^2}  576} \right)  100\left( {{x^2}  576} \right) = 0{/tex}
=> {tex}\left( {{x^2}  576} \right)\left( {{x^2}  100} \right) = 0{/tex}
=> {tex}{x^2}  576 = 0{/tex} and {tex}{x^2}  100 = 0{/tex}
=> {tex}{x^2} = 576{/tex} and {tex}{x^2} = 100{/tex}
=> {tex}x=24{/tex} m and {tex}x = 10{/tex} m
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ax + by = 1
bx + ay = (a+b)^2 / a2 + b2

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the age of the father is equal to sum of the ages of his 6 children. after 15 years twice the age of the father will be the sum of ages of his children. find age of father

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Let the age of father be x.
and the average age of one child be y.
Present age of father = sum of ages of 6 children
x = 6y ................................. (i)
After 15 years
Father's age = x + 15
Age of one child = y + 15
According to question
2(x + 15) = 6 (y + 15)
2x + 30 = 6y + 90
2 × 6y + 30 = 6y + 90 ....... [substitude x = 6y from (i)]
12y + 30 = 6y + 90
12y  6y = 90  30
6y = 60
{tex}\eqalign{ & \,\,\,\,\,y\,\, = \,\,{{60} \over 6}\,\, = \,\,10 \cr & \,\,x\,\, = \,\,6y\,\, = \,\,6\,\, \times \,\,10\,\, = \,\,60 \cr} {/tex}
Hence the age of father be = x = 60 years.
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From where can I get free Cbse sample papers class 10 (201718) for downloading, for all subjects.

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Log into mycbseguide.com and go in the Downloads section.
You can easily download the papers of all subjects according to your need. You may also download solutions.
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If one zero of the quadratic polynomial x^{2}+3x+k is 7,then find the value of k

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Let p(x) = x^{<font size="2">2 </font>}+ 3x + k
If one zero of the p(x) is 7, then
p(7) = 0
=> (7)^{2} + 3 x 7 + k = 0
=> 49 + 21 + k = 0
=> 70 + k = 0
=> k = 70
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Find the zeroes and verify the relationship between zeroes and coefficients:
4u^{2} + 8u
Please give step by step instructions and explain every step.
Please.

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4u^2 + 8 u +0 = 0
D=64
x = b + √ D÷2a,b √D÷2a
x = 0,2
2+0 = 8÷4
2= 2
2×0 = 0÷4
0=0
Hence,verified.
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Let p(u)=4u^{2}+0.u+8
=>4(u^{2}+2)
Therefore the req zeroes are 4 and √2
Verify it then
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