ANSWER QUESTIONS 22-24..
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Answers
Answer:
22) x²-(sum of zeros)x+(product of zeros)=0
x²-(-3/5)x+(-13/5)=0
x²+3x/5-13/5=0
5x²+3x-13/5=0
5x²+3x-13=0
required given polynomial is 5x²+3x-13
24)class marks=upper class +lower class/2
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Step-by-step explanation:
22)Given:-
Product and the sum of the zeroes are -13/5 and -3/5
To find:-
Write the polynomial ?
Solution:-
Product of the zeroes=-13/5
Sum of the zeroes=-3/5
The required polynomial=k[x²-(sum of the zeroes)x+(Product of the zeroes)]
=>k[x²-(-3/5)x+(-13/5)]
=>k[x²+3x/5-13/5]
=>k[(5x²+3x-13)]/5
If k=5 then the polynomial=5x²+3x-13
Answer:-
The required polynomial is 5x²+3x-13
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23)Given:-
Given points (3,1) and (1,5)
To find:-
What point on y-axis is equidistant from the given points.
Solution:-
Given points are (3,1) and (1,5)
Let the point on y-axis =(0,y)
(0,y) is the equidistant from (3,1) and (1,5)
(3,1)________(0,y)_________(1,5)
Distance between two points (x1, y1) and (x2, y2) is √{(x2-x1)²+(y2-y1)²} units
Distance between (3,1) and (0,y):-
Let (x1, y1)=(3,1)=>x1=3; y1=1
(x2, y2)=(0,y)=>x2=0 ; y2=y
Distance between them
=>√{(0-3)²+(y-1)²}
=>√{(-3)²+(y-1)²}
=>√(9+y²-2y+1)
=>√(y²-2y+10) units----------(1)
Distance between (0,y) and (1,5)
Let (x1, y1)=(0,y)=>x1=0 ; y1=y
(x2,y2)=(1,5)=> x2=1 ; y2=5
Distance between them
=>√{(1-0)²+(5-y)²}
=>√{1+(5-y)²}
=>√(1+25+y²-10y)
=>√(y²-10y+26) units----------(2)
According to the given problem
(1)=(2)
√(y²-2y+10)=√(y²-10y+26)
On squaring both sides then
=>(y²-2y+10)=(y²-10y+26)
=>-2y+10=-10y+26
=>-2y+10y=26-10
=>8y=16
=>y=16/8
=>y=2
Answer:-
The value of y=2
The required point=(0,2)
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24)
The formula for finding class mark of a class of the given data
=(Lower limit+Upper Limit)/2
Average of the lower and upper limits
Ex:-
class mark of 10-20 is (10+20)/2=30/2=15
Class mark of the class is also mid value of the class.
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