For this scenario, we can visualize that all points form a triangle. The three vertices are at the pitcher's mound, home plate, and the location where the outfielder catches the ball. We know two sides of the triangle and the angle that lies between these two sides.
<span>Using the cosine law, we can find the unknown third side. The formula to apply is:</span>
c^2 = a^2 + b^2 - 2ab cos θ
Where:
a = 60.5 ft
b = 195 ft
θ = 32°
Substituting the provided values results in:
c^2 = (60.5)^2 + (195)^2 - 2(60.5)(195) cos(32)
c^2 = 3660.25 + 38025 - 20009.7
c^2 = 21,675.56
c = 147.23 ft
<span>Thus, the distance the outfielder throws the ball towards home plate is approximately 147.23 ft.</span>
James will retain his original t toy cars along with half of (t+13) cars, resulting in a total of...
... t + (t + 13/2) = (3t + 13)/2.... cars after receiving a gift from Paul.
Response:
Step-by-step clarification:
Reply:
a) y-8 = (y₀-8), b) 2y -5 = (2y₀-5)
Clarification:
To address these equations, using direct integration is the simplest approach.
a) The equation provided is
dy / dt = -y + 8
dy / (y-8) = dt
We substitute variables
y-8 = u
dy = du
Substituting and integrating gives us
∫ du / u = ∫ dt
Ln (y-8) = t
Evaluating at the lower limits t = 0 for y = y₀
ln (y-8) - ln (y₀-8) = t-0
Simplifying the equation results in
ln (y-8 / y₀-8) = t
y-8 / y₀-8 =
y-8 = (y₀-8)
b) The equation here is
dy / dt = 2y -5
u = 2y -5
du = 2 dy
du / 2u = dt
Integrating now
½ Ln (2y-5) = t
Evaluating at limits gives
½ [ln (2y-5) - ln (2y₀-5)] = t
Ln (2y-5 / 2y₀-5) = 2t
2y -5 = (2y₀-5)
c) The equation here bears a strong resemblance to the previous one
u = 2y -10
du = 2 dy
∫ du / 2u = dt
ln (2y-10) = 2t
We evaluate this to get
ln (2y-10) –ln (2y₀-10) = 2t
2y-10 = (2y₀-10)
Answer:
Step-by-step explanation:
The equation provided is:
Applying the quadratic formula:
Based on the details given:
a = 2, b = -4, c = 7
Consequently:
=>
= 