Step-by-step explanation:
A consultant needs to make at least $800 this week. She earns $80 for each
1 answer:
You might be interested in
The statements and reasons why the given statements are true are presented in the following two column proofs;
Question 1
Given: 9·(x + 6) - 41 = 75
Statement Reason
S1. 9·(x + 6) - 41 = 75 R1. <u>Given</u>
S2. 9·(x + 6) = 116 R2. <u>Addition property</u>
S3. <u>9·x + 54 = 116 </u> R3. Distributive property
S4. 9·x = 62 R4. <u>Subtraction property</u>
S5.
R5. <u>Division property of equality</u>
Question 2.
Statement Reason
S1. m∠A + m∠B = m∠D R1. Given
S2. ∠C and ∠D form a Linear Pair R2. Definition of linear pair
S3. ∠C and ∠D are supplementary R3. <u>Linear pair ∠s are supplementary</u>
S4. <u>∠C + ∠D = 180° </u> R4. Definition of supplementary
S5. m∠C + m∠A + m∠B = 180° R5. <u>Substitution property</u>
Question 3.
Statement Reason
S1. ∠BDA ≅ ∠A R1. Given
S2. ∠BDA ≅ ∠CDE R2. <u>Vertical angle theorem</u>
S3. ∠CDE ≅ ∠A R3. Transitive property of congruency
S4. m∠CDE = m∠A R4. <u>Definition of congruency</u>
S5. <u>(13·x + 20)° = (14·x + 15)°</u> R5. Substitution Property of Equality
S6. 14·x° = 13·x + 5° R6. <u>Subtraction property</u>
S7. x = 5 R7. Subtraction property
Learn more here:
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