Isabella can afford a $410-per-month car payment, and she's interested in either a sedan, which costs $21,600, or a station wago
2 answers:
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Answer:
1) c ║ d by consecutive interior angles theorem
2) m∠3 + m∠6 = 180° by transitive property
3) ∠2 ≅ ∠5 by definition of congruency
4) t ║ v Corresponding angle theorem
5) ∠14 and ∠11 are supplementary Definition of supplementary angles
6) ∠8 and ∠9 are supplementary Consecutive interior angles theorem
Step-by-step explanation:
1) Statement Reason
m∠4 + m∠7 = 180° Given
m∠4 ≅ m∠6 Vertically opposite angles
m∠4 = m∠6 Definition of congruency
m∠6 + m∠7 = 180° Transitive property
m∠6 and m∠7 are supplementary Definition of supplementary angles
∴ c ║ d Consecutive interior angles theorem
2) Statement Reason
m∠3 = m∠8 Given
m∠8 + m∠6 = 180° Sum of angles on a straight line
∴ m∠3 + m∠6 = 180° Transitive property
3) Statement Reason
p ║ q Given
∠1 ≅ ∠5 Given
∠1 = ∠5 Definition of congruency
∠2 ≅ ∠1 Alternate interior angles theorem
∠2 = ∠1 Definition of congruency
∠2 = ∠5 Transitive property
∠2 ≅ ∠5 Definition of congruency.
4) Statement Reason
∠1 ≅ ∠5 Given
∠3 ≅ ∠4 Given
∠1 = ∠5 Definition of congruency
∠3 = ∠4 Definition of congruency
∠5 ≅ ∠4 Vertically opposite angles
∠5 = ∠4 Definition of congruency
∠5 = ∠3 Transitive property
∠1 = ∠3 Transitive property
∠1 ≅ ∠3 Definition of congruency.
t ║ v Corresponding angle theorem
5) Statement Reason
∠5 ≅ ∠16 Given
∠2 ≅ ∠4 Given
∠5 = ∠16 Definition of congruency
∠2 = ∠4 Definition of congruency
EF ║ GH Corresponding angle theorem
∠14 ≅ ∠16 Corresponding angles
∠14 = ∠16 Definition of congruency
∠5 = ∠14 Transitive property
∠5 + ∠11 = 180° Sum of angles on a straight line
∠14 + ∠11 = 180° Transitive property
∠14 and ∠11 are supplementary Definition of supplementary angles
6) Statement Reason
l ║ m Given
∠4 ≅ ∠7 Given
∠4 = ∠7 Definition of congruency
∠2 ≅ ∠7 Alternate angles
∠2 = ∠7 Definition of congruency
∠2 = ∠4 Transitive property
∠2 ≅ ∠4 Definition of congruency
∠2 and ∠4 are corresponding angles Definition
DA ║ EB Corresponding angle theorem
∠8 and ∠9 are consecutive interior angles Definition
∠8 and ∠9 are supplementary Consecutive interior angles theorem.
Answer: He cannot see a ship that is 4.5 miles away.
Step-by-step explanation:
Since we have given that
Here, h is the eye level above the water
d is the distance away.
If h = 12 feet
We will get :
Hence, it is 10.39 feet away when eye level is 12 feet above the water.
Therefore , he cannot see a ship that is 4.5 miles away.
The attached figure represents the relation between ω (rpm) and t (seconds)
To find the blade's angular position in radians ⇒ ω will be converted from (rpm) to (rad/s)
ω = 250 (rpm) = 250 * (2π/60) = (25/3)π rad/s
ω = 100 (rpm) = 100 * (2π/60) = (10/3)π rad/s
and from the figure it is clear that the operation is at constant speed but with variable levels
⇒ ω = dθ/dt ⇒ dθ = ω dt
∴ θ = ∫₀²⁰ ω dt
while ω is not fixed from (t = 0) to (t =20)
the integral will divided to 3 integrals as follow;
ω = 0 from t = 0 to t = 5
ω = 250 (rpm) = (25/3)π from t = 5 to t = 15
ω = 100 (rpm) = (10/3)π from t = 15 to t = 20
∴ θ = ∫₀⁵ (0) dt + ∫₅¹⁵ (25/3)π dt + ∫₁₅²⁰ (10/3)π dt
the first integral = 0
the second integral = (25/3)π t = (25/3)π (15-5) = (250/3)π
the third integral = (10/3)π t = (25/3)π (20-15) = (50/3)π
∴ θ = 0 + (250/3)π + (50/3)π = 100 π
while the complete revolution = 2π
so instantaneously at t = 20
∴ θ = 100 π - 50 * 2 π = 0 rad
Which mean:
the blade will be at zero position making no of revolution = (100π)/(2π) = 50
To find the blade's angular position in radians ⇒ ω will be converted from (rpm) to (rad/s)
ω = 250 (rpm) = 250 * (2π/60) = (25/3)π rad/s
ω = 100 (rpm) = 100 * (2π/60) = (10/3)π rad/s
and from the figure it is clear that the operation is at constant speed but with variable levels
⇒ ω = dθ/dt ⇒ dθ = ω dt
∴ θ = ∫₀²⁰ ω dt
while ω is not fixed from (t = 0) to (t =20)
the integral will divided to 3 integrals as follow;
ω = 0 from t = 0 to t = 5
ω = 250 (rpm) = (25/3)π from t = 5 to t = 15
ω = 100 (rpm) = (10/3)π from t = 15 to t = 20
∴ θ = ∫₀⁵ (0) dt + ∫₅¹⁵ (25/3)π dt + ∫₁₅²⁰ (10/3)π dt
the first integral = 0
the second integral = (25/3)π t = (25/3)π (15-5) = (250/3)π
the third integral = (10/3)π t = (25/3)π (20-15) = (50/3)π
∴ θ = 0 + (250/3)π + (50/3)π = 100 π
while the complete revolution = 2π
so instantaneously at t = 20
∴ θ = 100 π - 50 * 2 π = 0 rad
Which mean:
the blade will be at zero position making no of revolution = (100π)/(2π) = 50