∠CAE = 120°
∠CAD = 60°
∠BAE = 180°
∠DEC = 30°
We start out with the fact that points C and D split the semicircle into 3 sections. This means that ∠BAC, ∠CAD and ∠DAE are all 60° (180/3 = 60).
Since it forms a straight line, ∠BAE is 180°.
Since it is formed by ∠CAD and ∠DAE, ∠CAE = 60+60 = 120°.
We know that an inscribed angle is 1/2 of the corresponding arc; since CD is 1/3 of the circle, it is 1/3(180) = 60; and this means that ∠DEC = 30°.
Answer:
<em>after 4seconds</em>
Step-by-step explanation:
Given the height, h , in feet, of the football above the ground after t seconds expressed by h ( t ) = − 8 t^2 + 32 t, the height of the ball on the ground is 0feet.
Substitute h(t) = 0 into the expression and calculate t;
h ( t ) = − 8 t^2 + 32 t
0 = − 8 t^2 + 32 t
8t² = 32t
8t = 32
Divide both sides by 8
8t/8 = 32/8
<em>t = 4s</em>
<em>Hence the football hits the ground after 4seconds</em>
Given:
In triangle GHJ, ∠G = 110°, ∠J = 40° and ∠H = 30°.
To find:
The answer to complete the given statements.
Solution:
According to triangle side and angle relationship, largest angle has longest opposite side and smallest angle has shortest opposite side.
∠G = 110°, ∠J = 40° and ∠H = 30°.
Here, ∠G > ∠J = 40° and ∠G > ∠H. So, ∠G is the Largest angle.
Since angle G is largest angle, the opposite side, JH, is longest.
Clearly,
110° > 40° > 30°.
∠G > ∠J > ∠H
Using triangle side and angle relationship, we get
JH > GH > GJ
The order of the side lengths from longest to shortest is JH, GH ahd GJ.
The x-coordinate of the axis os symmetry of a quadratic is the x-coorfinate of its vertex. To find the y-coordinate of the vertex here we would substitute -5 for x in:
The coordinates of the vertex are (-5,41)
Answer:
Options (3) and (6)
Step-by-step explanation:
ΔABC is a dilated using a scale factor of 
to produce image triangle ΔA'B'C'.
Since, dilation is a rigid transformation,
Angles of both the triangles will be unchanged or congruent.
m∠A = m∠A' and m∠B = m∠B'
Since, sides of ΔA'B'C' = of the sides of ΔABC
Area of ΔA'B'C' = 
Area of ΔABC > Area of ΔA'B'C'
Since, angles of ΔABC and ΔA'B'C' are congruent, both the triangles will be similar.
ΔABC ~ ΔA'B'C'
Therefore, Option (3) and Option (6) are the correct options.
