Answer:
(A)6
Step-by-step explanation:
Given the quadratic expression:
We factorize:
Therefore, the missing number that will complete the factorization is 6.
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Answer:
<h3>Q cuts the diagonal PA into 2 equal halves, since the diagonals of rhombus meet at right angles.</h3><h3>The value of x is 8.</h3>Step-by-step explanation:
Given that Quadrilateral CAMP below is a rhombus. the length PQ is (x+2) units, and the length of QA is (3x-14) units
From the given Q is the middle point, which cut the diagonal PA into 2 equal halves.
By the definition of rhombus, diagonals meet at right angles.
Implies that PQ = QA
x+2 = 3x - 14
x-3x=-14-2
-2x=-16
2x = 16
dividing by 2 on both sides, we will get,
The line segment
( since x=8)
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
Answer:
Approximately, 159 men weighs more than 165 pounds and 159 men weighs less than 135 pounds.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 150 pounds
Standard Deviation, σ = 15
We are given that the distribution of weights of 1000 men is a bell shaped distribution that is a normal distribution.
Formula:
P( men weighing more than 165 pounds)
P(x > 165)
Calculation the value from standard normal z table, we have,
Approximately, 159 men weighs more than 165 pounds.
P(men weighing less than 135 pounds)
P(x < 135)
Calculation the value from standard normal z table, we have,
Approximately, 159 men weighs less than 135 pounds.