Answer:
Please find attached a drawing of the triangles ΔRST and EFG showing the angles
The angle on ΔEFG that would prove the triangles are similar is ∠F = 25°
Step-by-step explanation:
In order to prove that two triangles are similar, two known angles of each the triangles need to be shown to be equal
Given that triangle ∠R and ∠S of triangle ΔRST are 95° and 25°, respectively, and that ∠E of ΔEFG is given as 90°, then the corresponding angle on ΔEFG to angle ∠S = 25° which is ∠F should also be 25°
Therefore, the angle on ΔEFG that would prove the triangles are similar is ∠F = 25°.
Answer:
46
Step-by-step explanation: .23 (200) = 46
The height at t seconds after launch is
s(t) = - 16t² + V₀t
where V₀ = initial launch velocity.
Part a:
When s = 192 ft, and V₀ = 112 ft/s, then
-16t² + 112t = 192
16t² - 112t + 192 = 0
t² - 7t + 12 = 0
(t - 3)(t - 4) = 0
t = 3 s, or t = 4 s
The projectile reaches a height of 192 ft at 3 s on the way up, and at 4 s on the way down.
Part b:
When the projectile reaches the ground, s = 0.
Therefore
-16t² + 112t = 0
-16t(t - 7) = 0
t = 0 or t = 7 s
When t=0, the projectile is launched.
When t = 7 s, the projectile returns to the ground.
Answer: 7 s
Answer:
let that man's age be x :
subtract 300 from both sides:
apply the division rule of equality:
<u>The</u><u> </u><u>man</u><u> </u><u>is</u><u> </u><u>9</u><u>8</u><u> </u><u>years</u><u> </u>