21 Solve the linear equations 2x-y-5 and 3x+2y-1 using substitution method
22. Check whether - 150 is a term of the AP: 11.8.5.2
23. Ifx +2x+6 are in GP Find value
24 Find the point on x axis which is equistance from the points (2-5) and (-29)
25. Find the area of the triangle whose vertices are (1-1).(-4,6) and (-3.-5)
1 + sine
26. Ir cote- then evaluate the value of -
COS
27 IFAB and Care the interior angles of the triangle ABC then show that sin
28. If the length of the shadow of a 15 meter high pole is 53 meters on the ground at 701
clock in the morning. Find an angle of elevation
Answers
Answer:
Step-by-step explanation:
Given: the linear equations 2x-y-5 = 0 and 3x+2y-1 = 0
To find : Value of x & y using substitution method
Solution:
2x-y-5 = 0
=> y =2x - 5
3x+2y-1 = 0
putting y = 2x - 5
=> 3x + 2(2x - 5) - 1 = 0
=> 3x + 4x- 10 - 1 = 0
=> 7x = 11
=> x = 11/7
y = 2x - 5
=> y = 2(11/7) - 5
=> y = 22/7 - 5
=> y = -13/7
( 11/7 , - 13/7)
AP : 11 , 8 , 5 , 2
=> a = 11 d = - 3
-150 = 11 + (n-1)(-3)
=> (n - 1) = 161/3
n not an integer
so -150 is not a term of AP 11 , 8 , 5 , 2
Point on x axis = ( x , 0)
equi distance from the points (2-5) and (-2, 9)
=> (x - 2)² + 5² = (x + 2)² + 9²
=> x² - 4x + 4 + 25 = x² + 4x + 4 + 81
=> 8x = -56
=> x = -7
( - 7 , 0) is equi distance from the points (2-5) and (-2, 9)
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