180° = 70° + (2x-10°) + (x+30°)
180° - 70° = (2x-10°) + (x+30°)
110° = 3x + 20°
110° - 20° = 3x
90° = 3x
x = 30°
A is 2x - 10°
A = 2(30) - 10
A = 60 - 10
A = 50°
Answer:
Given: Consider a triangle ABC in which AD is median drawn from vertex A.
To prove: AB + AC > AD
Proof: In Δ A B D
AB + B D> AD ⇒[In a triangle sum of lengths of two sides is greater than the third side] .................(1)
In Δ A CD
AC + DC > AD [In a triangle sum of lengths of two sides is greater than the third side] .................(2)
Adding (1) and (2)
AB + AC+ B D + D C > B D + DC
A B+ A C+ B C > 2 B D .................(3)
Also , Considering Δ AB C
AB + B C > B C ⇒[In a triangle sum of lengths of two sides is greater than the third side]
⇒ AB + B C - B C >0 ........................(4)
Adding (3) and (4)
A B+ B C+B C+ AB +A C- B C > 2 A D
⇒2 AB + 2 A C> 2 A D
Dividing both side of inequality by 2, we get
A B+ A C> A D
Hence proved.
He got a discount of 32%, and saved 64$,
let the total amount without discount be x
according to the formula
32$=64×100/x
x=64×100/32
x=200$
Answer:
+ 12
Step-by-step explanation:
8 - (- 4) = 8 + 4 = 12
Answer:
(x-y-1)(x^2+(x+y)+1)
Step-by-step explanation: