Answer:
4
Step-by-step explanation:
use the Pythagoras' theorem
formula for an ABC triangle( right angled)
AB is height
BC is hypotenuse
CA is length
=
+
in your case
AB is 7
BC is 8
CA is unknown so we write x
=
+
64=49+
64-49=
15=
=x
3.87=x
4=x
Answer:
Given
Step-by-step explanation:
Given that: △RST ~ △VWX, TU is the altitude of △RST, and XY is the altitude of △VWX.
Comparing △RST and △VWX;
TU ~ XY (given altitudes of the triangles)
<TUS = <XYW (all right angles are congruent)
<UTS ≅ <YXW (angle property of similar triangles)
Thus;
ΔTUS ≅ ΔXYW (congruent property of similar triangles)
<UTS + <TUS + < UST = <YXW + <XTW + <XWY = (sum of angles in a triangle)
Therefore by Angle-Angle-Side (AAS), △RST ~ △VWX
So that:
=
(corresponding side length proportion)
Answer:
The baseball reached 9 feet high above the ground
Step-by-step explanation:
The given quadratic function representing the position of the baseball above ground is p(t) = 1/2·g·t² + v₀·t + p₀
Where;
t ≥ 0
g = -32 ft./sec²
v₀ = The initial velocity
p₀ = The initial position
Given that when the ball is thrown, we have;
The initial, straight up, velocity, v₀ = 16 ft./sec
The initial position, p₀ = 5 ft.
Substituting the above values in the quadratic function representing the position of the baseball above ground, we have;
p(t) = 1/2·(-32)·t² + 16·t + 5 = 16·t - 16·t² + 5
At the maximum point, the rate of change of the height with time = 0, therefore;
dp(t)/dt = 0 = d(16·t - 16·t² + 5)/dt = 16 - 32·t
16 - 32·t = 0
16 = 32·t
t = 16/32 = 0.5 seconds
Therefore, the time takes to reach the maximum height = 0.5 seconds
The height (maximum) reached in 0.5 seconds is given as follows;
h(t) = 16·t - 16·t² + 5, from which we have;
h(0.5) = 16 × 0.5 - 16 × (0.5)² + 5 = 9
Therefore, the height baseball reached = 9 ft. above ground
