Both of these problems will be solved in a similar way, but with different numbers. First, we set up an equation with the values given. Then, we solve. Lastly, we plug into the original expressions to solve for the angles.
[23] ABD = 42°, DBC = 35°
(4x - 2) + (3x + 2) = 77°
4x+ 3x + 2 - 2 = 77°
4x+ 3x= 77°
7x= 77°
x= 11°
-
ABD = (4x - 2) = (4(11°) - 2) = 44° - 2 = 42°
DBC = (3x + 2) = (3(11°) + 2) = 33° + 2 = 35°
[24] ABD = 62°, DBC = 78°
(4x - 8) + (4x + 8) = 140°
4x + 4x + 8 - 8 = 140°
4x + 4x = 140°
8x = 140°
8x = 140°
x = 17.5°
-
ABD = (4x - 8) = (4(17.5°) - 8) = 70° - 8° = 62°
DBC =(4x + 8) = (4(17.5°) + 8) = 70° + 8° = 78°
Answer:
L (1, - 7 )
Step-by-step explanation:
give endpoints (x₁, y₁ ( and (x₂, y₂ ) then the midpoint is
( ,
)
here (x₁, y₁ ) = K (9, - 7 ) and (x₂, y₂ ) = L (x, y )
use the midpoint formula and equate to corresponding coordinates of M
= 5 ( multiply both sides by 2 to clear the fraction )
9 + x = 10 ( subtract 9 from both sides )
x = 1
and
= - 7 ( multiply both sides by 2 )
- 7 + y = - 14 ( add 7 to both sides )
y = - 7
Then L = (1, - 7 )
Answer:
t = 2
Step-by-step explanation:
Notice that this expression for the projectile's path is that of a quadratic function with negative leading term. The graph of it therefore consists of a parabola with the branches pointing down (due to he negative leading coefficient). Therefore, the maximum of such parabola will reside at its vertex.
Recall that the formula for the position of the vertex in a general parabolic function of the form: , is given by the expression:
In our case, the variable "x" is in fact "t", the leading coefficient () is -5, and the coefficient for the linear term (
) is 20.
Therefore, the maximum of the path will be when