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Answer eight ----- given that
( x1 + x2 )/2 = 6 then
x1 + x2 = 12
now given that
(x1 + x2 + x3)/3 = 7
x1 + x2 + x3 = 21
and we have value of x1 + x2 , put in above equation ,then
12+ x3 = 21
x3 = 9
Answer ten :
first no. - 6 teachers has age less than 30
second no. - 20 teacher are above age of 30
( x1 + x2 )/2 = 6 then
x1 + x2 = 12
now given that
(x1 + x2 + x3)/3 = 7
x1 + x2 + x3 = 21
and we have value of x1 + x2 , put in above equation ,then
12+ x3 = 21
x3 = 9
Answer ten :
first no. - 6 teachers has age less than 30
second no. - 20 teacher are above age of 30
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1) let O be the center of the circle and AB is the chord. Join OA ,OB to form triangle AOB. Draw a perpendicular bisector from the center to AB at a point D.Therefore AD=BD. By pythagoreous theorem
OA=17cm , OD =8cm, AD =?
Substitute this values in the equation,
ad= 15
length of chord=AD+BD
Therefore length of chord=15+15=30
OA=17cm , OD =8cm, AD =?
Substitute this values in the equation,
ad= 15
length of chord=AD+BD
Therefore length of chord=15+15=30
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