Solve the system. x+3y=1 -2x+7=-23
1 answer:
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Answer:
3a) The value of x = 56
3b) The measure of ∠ H T M = 90°
3c) The radius of the circle = 53
Step-by-step explanation:
3a) ∵ A F is a tangent to the circle O at point F
∵ Secant AH intersects circle O at point T
∴ (A F)² = (A T)(A H)
∴ 7( x + 7) = (21)² ⇒ ÷ 7
∴ x + 7 = 63
∴ x = 63 - 7 = 56
3b) ∵ HM is a diameter
∴ The measure of the arc HM = 180° ⇒ semi-circle
∵ ∠ H T M is inscribed angle subtended by the arc HM
∴ m ∠ H T M = half the measure of arc HM
∴ m ∠ H T M = 180° ÷ 2 = 90°
3c) ∵ Δ H T M is a right angle triangle at T
∴ (H M)² = (M T)² + (H T)² ⇒ Pythagorean theorem
∴ (H M)² = (90)² + (56)²
∴ (H M)² = 11236
∴ HM = = 106
∴ OM = 106 ÷ 2 = 53
∵ OM is the radius of the circle O
∴ The radius = 53
Answer:
The maximum velocity of the object in the time interval [0, 4] is 10 ft/s.
Step-by-step explanation:
Given : The position of an object along a vertical line is given by , where s is measured in feet and t is measured in seconds.
To find : What is the maximum velocity of the object in the time interval [0, 4]?
Solution :
The velocity is rate of change of distance w.r.t time.
Distance in terms of t is given by,
Derivate w.r.t. time,
It is a quadratic function so its maximum is at vertex of the function.
The x point of the function is given by,
Where, a=-3, b=6 and c=7
As 1 lie between interval [0,4]
Substitute t=1 in the function,
Th maximum velocity is 10 ft/s.
Therefore, the maximum velocity of the object in the time interval [0, 4] is 10 ft/s.