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Sagot :
Two large and 1 small pumps can fill a swimming pool in 4 hours. One large and 3 small pumps can also fill the same swimming pool in 4 hours. How many hours will it take 4 large and 4 small pumps to fill the swimming pool.(We assume that all large pumps are similar and all small pumps are also similar.)
Find all sides of a right triangle whose perimeter is equal to 60 cm and its area is equal to 150 square cm.
A circle of center (-3 , -2) passes through the points (0 , -6) and (a , 0). Find a.
Find the equation of the tangent at ( 0 , 2) to the circle with equation
(x + 2)2 + (y + 1)2 = 13
An examination consists of three parts. In part A, a student must answer 2 of 3 questions. In part B, a student must answer 6 of 8 questions and in part C, a student must answer all questions. How many choices of questions does the student have?
Solve for x
x2 - 3|x - 2| - 4x = - 6
The right triangle ABC shown below is inscribed inside a parabola. Point B is also the maximum point of the parabola (vertex) and point C is the x intercept of the parabola. If the equation of the parabola is given by y = -x2 + 4x + C, find C so that the area of the triangle ABC is equal to 32 square units.
∫ (x³cos(x/2)+(1/2))√(4-x²) dx [-2,2]
Distribute
∫ x³√(4-x²)cos(x/2) + (1/2)√(4-x²) dx [-2,2]
Now
Determine the function of
f(x) = x³√(4-x²)cos(x/2)
Is an odd or a function
f(-x) = -x³√(4-x²)cos(x/2)
Therefore the function is odd so automatically the answer for
∫ x³√(4-x²)cos(x/2) dx [-2,2] = 0
While for
(1/2)∫ √(4-x²) dx[-2,2]
f(-x) = √(4-x²)
So the function is even therefore
(1/2)•2 ∫ √(4-x²) dx [0,2]
∫ √(4-x²) dx [0,2]
x = 2sinθ
dx = 2cosθ dθ
2 ∫ √(4-4sin²θ)cosθ dθ
4 ∫ cos²θ dθ
2 ∫ 1+cos(2θ) dθ
2(θ+(1/2)sin(2θ))|[0,2]
2θ+sin(2θ)|[0,2]
2arcsin(x/2)+(1/2)x√(4-x²)|[0,2]
2arcsin(1)+0
= 2(π/2)
= π
✮
#ᴄᴀʀʀʏᴏɴʟᴇᴀʀɴɪɴɢ
Find all sides of a right triangle whose perimeter is equal to 60 cm and its area is equal to 150 square cm.
A circle of center (-3 , -2) passes through the points (0 , -6) and (a , 0). Find a.
Find the equation of the tangent at ( 0 , 2) to the circle with equation
(x + 2)2 + (y + 1)2 = 13
An examination consists of three parts. In part A, a student must answer 2 of 3 questions. In part B, a student must answer 6 of 8 questions and in part C, a student must answer all questions. How many choices of questions does the student have?
Solve for x
x2 - 3|x - 2| - 4x = - 6
The right triangle ABC shown below is inscribed inside a parabola. Point B is also the maximum point of the parabola (vertex) and point C is the x intercept of the parabola. If the equation of the parabola is given by y = -x2 + 4x + C, find C so that the area of the triangle ABC is equal to 32 square units.
∫ (x³cos(x/2)+(1/2))√(4-x²) dx [-2,2]
Distribute
∫ x³√(4-x²)cos(x/2) + (1/2)√(4-x²) dx [-2,2]
Now
Determine the function of
f(x) = x³√(4-x²)cos(x/2)
Is an odd or a function
f(-x) = -x³√(4-x²)cos(x/2)
Therefore the function is odd so automatically the answer for
∫ x³√(4-x²)cos(x/2) dx [-2,2] = 0
While for
(1/2)∫ √(4-x²) dx[-2,2]
f(-x) = √(4-x²)
So the function is even therefore
(1/2)•2 ∫ √(4-x²) dx [0,2]
∫ √(4-x²) dx [0,2]
x = 2sinθ
dx = 2cosθ dθ
2 ∫ √(4-4sin²θ)cosθ dθ
4 ∫ cos²θ dθ
2 ∫ 1+cos(2θ) dθ
2(θ+(1/2)sin(2θ))|[0,2]
2θ+sin(2θ)|[0,2]
2arcsin(x/2)+(1/2)x√(4-x²)|[0,2]
2arcsin(1)+0
= 2(π/2)
= π
✮
#ᴄᴀʀʀʏᴏɴʟᴇᴀʀɴɪɴɢ
1.) 5/6
2.) 1/2
3.) 14/5
4.) 4/5
5.) 2/5
6.)24 1/5
7.) 2/15
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